If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Exponential Functions and Logarithmic Functions

Samuel Dominic Chukwuemeka (SamDom For Peace) For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB, NZQA, and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve each question.
Show all work.

(1.) Identify whether the statement represents an exponential function.
For each training session, a personal trainer charges his clients $\$5$ less than the previous training session.


Let us represent this information in a table.
Let the fee for the first training session be $p$
Training Session Fee ($\$$)
$1$ $p$
$2$ $p - 5$
$3$ $(p - 5) - 5$
$p - 5 - 5$
$p - 10$
$p - 2(5)$
$4$ $(p - 10) - 5$
$p - 10 - 5$
$p - 15$
$p - 3(5)$
$p - (4 - 1)(5)$
$x$ $p - (x - 1)(5)$
$p - 5(x - 1)$
$p - 5x + 5$
This is a linear function
It is not an exponential function.

An exponential function has:
(1.) a base that is a constant
(2.) an exponent that is the independent variable.
Examples: $y = 3^x$, $y = e^x$ among others.
(2.)

(3.) All the graphs shown below have the form: $f(x) = ab^x$
Question 3

(a.) Which graph has the smallest value of $a$?
(b.) Which graph has the largest value of $a$?
(c.) Which graph has the smallest value of $b$?
(d.) Which graph has the largest value of $b$?


(a.)
The graph with the smallest value of $a$ is the graph with the least $y-intercept$
This implies that when $x = 0$, the graph that has the least value of $y$ (the least value of $y$ on the $y-axis$) is the graph with the smallest value of $a$
That graph is Graph F

(b.)
The graph with the largest value of $a$ is the graph with the greatest $y-intercept$
This implies that when $x = 0$, the graph that has the greatest value of $y$ (the greatest value of $y$ on the $y-axis$) is the graph with the largest value of $a$
That graph is Graph C

(c.)
The graph with the samllest value of $b$ is the graph that:

(1.) decreases from left to right.
It is a decreasing function which signifies an exponential decay.
and
(2.) has the steepest slope.

The first condition eliminates Graphs D, E, F because those graphs are increasing from left to right and their bases are greater than 1

We are left with Graphs A, B, C
These graphs are decreasing from left to right. Their bases are between 0 and 1
Looking at these graphs, the one with the steepest slope is Graph C
Therefore, Graph C has the smallest value for the base, $b$

Let us verify with some graph examples:
Answer 3


(d.)
The graph with the largest value of $b$ is the graph that:

(1.) increases from left to right.
It is an increasing function which signifies an exponential growth.
and
(2.) has the steepest slope.

The first condition eliminates Graphs A, B, C because those graphs are decreasing from left to right and their bases are between 0 than 1

We are left with Graphs D, E, F
These graphs are increasing from left to right. Their bases are greater than 1
Looking at these graphs, the one with the steepest slope is Graph D
Therefore, Graph D has the largest value for the base, $b$

Let us verify with some graph examples:
Answer 3
(4.) $x^7 = 128$


First Method - By Exponents $ x^7 = 128 \\[2ex] x^7 = 2^7 \\[2ex] Exponents\: are\: the\: same \\[2ex] Equate\: the\: bases \\[2ex] x = 2 $
Second Method - By Logarithms $ x^7 = 128 \\[2ex] Introduce\: \log_2\: to\: both\: sides \\[2ex] \log_2{x^7} = \log_2{128} \\[2ex] \log_2{x^7} = 7\log_2{x} ...Law\: 5...Log \\[2ex] \log_2{128} = \log_2{2^7} = 7\log_2{2} ... Law\: 5...Log \\[2ex] \log_2{2} = 1...Law\: 4...Log \\[2ex] \log_2{128} = 7 * 1 = 7 \\[2ex] \rightarrow 7\log_2{x} = 7 \\[2ex] Divide\: both\: sides\: by\: 7 \\[2ex] \log_2{x} = 1 \\[2ex] 1 = \log_2{2} ...Law\: 4...Log \\[2ex] \log_2{x} = \log_2{2} \\[2ex] LogBase\: is\: the\: same \\[2ex] Equate\: the\: terms \\[2ex] x = 2 $
Check

LHS
$ x^7 \\[2ex] 2^{7} \\[2ex] 128 $

RHS
$ 128 $

(5.) Select the correct graph of the pair of functions on the same axis.

$ f(x) = e^x \;\;\;and\;\;\; g(x) = \ln x \\[3ex] $ The function $f(x)$ is the solid blue curve and $g(x)$ is the dashed red curve.
Question 5-1
Question 5-2


First:
$f(x)$ and $g(x)$ are inverses of each other.
This implies that the domain of $f(x)$ is the range of $g(x)$
and
the range of $f(x)$ is the domain of $g(x)$

$ f(x) \\[3ex] Domain:\;\; (-\infty, \infty) \\[3ex] Range:\;\; (0, \infty) \\[5ex] $ This eliminates Graphs A and B because the domain for Graphs A and B is $(0, \infty)$

$ g(x) \\[3ex] Domain:\;\; (0, \infty) \\[3ex] Range:\;\; (-\infty, \infty) \\[5ex] $ We are left with Graphs C and D

Second:
Any point say $(c, d)$ on the graph of $f(x)$ is the point $(d, c)$ on the graph of $g(x)$
This is because $g(x)$ is the inverse of $f(x)$
Any graph that does not have the point $(c, d)$ on the graph of $f(x)$ and the point $(d, c)$ on the graph of $g(x)$ is not the correct graph.

$f(x) = e^x$ $g(x) = \ln x$
$x$ $y$ $x$ $y$
$0$ $1$ $1$ $0$

The $y-intercept$ of $f(x)$ is the $x-intercept$ of $g(x)$
Both Graphs C and D satisfy this.
Let us try another point.

$f(x) = e^x$ $g(x) = \ln x$
$x$ $y$ $x$ $y$
$1$ $e$ ($\approx 2.718$) $e$ ($\approx 2.718$) $1$

Looking at both graphs, Graph C is the only graph to satisfy these points for $f(x)$ and $g(x)$
For Graph D, for $g(x)$; when $x = 2.718$, $y \lt 1$
Therefore, Graph C is the correct option.
(6.) $x$${\dfrac{1}{4}}$ = $2$


First Method - By Exponents $ x^{\dfrac{1}{4}} = 2 \\[2ex] Multiply\: both\: exponents\: by\: 4 \\[2ex] x^{\dfrac{1}{4} * 4} = 2^4 \\[2ex] x = 16 $
Second Method - By Logarithms $ x^{\dfrac{1}{4}} = 2 \\[2ex] Introduce\: \log_2\: to\: both\: sides \\[2ex] \rightarrow \log_2{x^{\dfrac{1}{4}}} = \log_2{2} \\[2ex] \log_2{x^{\dfrac{1}{4}}} = \dfrac{1}{4}\log_2{x} ...Law\: 5...Log \\[2ex] \log_2{2} = 1...Law\: 4...Log \\[2ex] \rightarrow \dfrac{1}{4}\log_2{x} = 1 \\[2ex] Multiply\: both\: sides\: by\: 4 \\[2ex] \log_2{x} = 4 \\[2ex] 4 = \log_2{2^4} ...Laws\: 4\: and\: 5...Log \\[2ex] \rightarrow \log_2{x} = \log_2{2^4} \\[2ex] LogBase\: is\: the\: same \\[2ex] Equate\: the\: terms \\[2ex] x = 2^4 \\[2ex] x = 16 $
Check

LHS
$ x^{\dfrac{1}{4}} \\[3ex] 16^{\dfrac{1}{4}} \\[3ex] \sqrt[4]{16} \\[3ex] 2 $

RHS
$ 2 $

(7.) $9^{x^{2}} * 3^{3x} = 9$


First Method - By Exponents $ 9^{x^{2}} * 3^{3x} = 9 \\[3ex] 9 = 3^2 \\[3ex] 9^{x^{2}} = 3^{{2} ({x^{2}})} \\[3ex] 3^{{2} ({x^{2}})} = 3^{2x^{2}} ...Law\: 5...Exp \\[3ex] \rightarrow 3^{2x^{2}} * 3^{3x} = 3^{2x^2 + 3x} ...Law\: 1...Exp \\[3ex] 3^{2x^2 + 3x} = 3^2 \\[3ex] Same\: base \\[2ex] Equate\: the\: exponents \\[2ex] 2x^2 + 3x = 2 \\[3ex] 2x^2 + 3x - 2 = 0 \\[3ex] 2x^2 + 4x - x - 2 = 0 \\[3ex] 2x(x + 2) - 1(x + 2) = 0 \\[3ex] x + 2 = 0\:\:\: OR\:\:\: 2x - 1 = 0 \\[3ex] x = -2\:\:\: OR\:\:\: 2x = 1 \\[3ex] x = -2\:\:\: OR\:\:\: x = \dfrac{1}{2} \\ $
Second Method - By Logarithms $ 9^{x^{2}} * 3^{3x} = 9 \\[3ex] Introduce\: \log_3\: to\: both\: sides \\[3ex] \rightarrow \log_3{(9^{x^{2}} * 3^{3x})} = \log_3{9} \\[3ex] \log_3{(9^{x^{2}} * 3^{3x})} = \log_3{9^{x^{2}}} + \log_3{3^{3x}} ...Law\: 1...Log \\[3ex] \log_3{9^{x^{2}}} = x^2\log_3{9} ... Law\: 5...Log \\[3ex] \log_3{9} = \log_3{3^2} = 2\log_3{3} ...Law\: 5...Log \\[3ex] \log_3{3} = 1...Law\: 4...Log \\[3ex] 2\log_3{3} = 2 * 1 = 2 \\[3ex] x^2\log_3{9} = x^2 * 2 * 1 = 2x^2 \\[3ex] \log_3{3^{x}} = 3x\log_3{3} ... Law\: 5...Log \\[3ex] 3x\log_3{3} = 3x * 1 = 3x \\[3ex] \rightarrow x^2\log_3{9} + 3x\log_3{3} = 2\log_3{3} \\[3ex] 2x^2 + 3x = 2 \\[3ex] 2x^2 + 3x - 2 = 0 \\[3ex] 2x^2 + 4x - x - 2 = 0 \\[3ex] 2x(x + 2) - 1(x + 2) = 0 \\[3ex] x + 2 = 0\:\:\: OR\:\:\: 2x - 1 = 0 \\[3ex] x = -2\:\:\: OR\:\:\: 2x = 1 \\[3ex] x = -2\:\:\: OR\:\:\: x = \dfrac{1}{2} $
Check

LHS
$ 9^{x^{2}} * 3^{3x} \\[3ex] x = -2 \\[3ex] = 9^{(-2)^{2}} * 3^{3(-2)} \\[3ex] = 9^{4} * 3^{-6} \\[3ex] = 3^{2(4)} * 3^{-6} \\[3ex] 3^{2(4)} = 3^{2 * 4} = 3^8 ...Law\: 5...Exp \\[3ex] = 3^8 * 3^{-6} \\[3ex] = 3^{8 + (-6)} ...Law\: 1...Exp \\[3ex] = 3^{8 - 6} \\[3ex] = 3^2 \\[3ex] = 9 $


LHS
$ 9^{x^{2}} * 3^{3x} \\[3ex] x = \dfrac{1}{2} \\[3ex] = 9^{\left(\dfrac{1}{2}\right)^{2}} * 3^{3\left(\dfrac{1}{2}\right)} \\[3ex] = 9^{\dfrac{1}{4}} * 3^{\dfrac{3}{2}} \\[3ex] = 3^{2\left(\dfrac{1}{4}\right)} * 3^{\dfrac{3}{2}} \\[3ex] 3^{2\left(\dfrac{1}{4}\right)} = 3^{2 * \dfrac{1}{4}} = 3^{\dfrac{1}{2}} ...Law\: 5...Exp \\[3ex] = 3^{\dfrac{1}{2}} * 3^{\dfrac{3}{2}} \\[3ex] = 3^{\left(\dfrac{1}{2} + \dfrac{3}{2}\right)} ...Law\: 1...Exp \\[3ex] = 3^{\left(\dfrac{1 + 3}{2}\right)} \\[3ex] = 3^{\dfrac{4}{2}} \\[3ex] = 3^2 \\[3ex] = 9 $

RHS
$ 9 $

(8.) $\left(\sqrt[5]{7}\right)$$1 - 4x$ = $7^{x^2}$


First Method - By Exponents
$\left(\sqrt[5]{7}\right)$$1 - 4x$ = $7^{x^2}$

$\sqrt[5]{7}$ = $7$$\dfrac{1}{5}$ $...Law\: 7...Exp$

$\left(\sqrt[5]{7}\right)$$1 - 4x$ = $7$$\left(\dfrac{1}{5}\right) * (1 - 4x)$

$7$$\left(\dfrac{1}{5}\right) * (1 - 4x)$ = $7$$\left(\dfrac{1 - 4x}{5}\right)$ $...Law\: 7...Exp$

$ \rightarrow 7^{\left(\dfrac{1 - 4x}{5}\right)} = 7^{x^2} \\[3ex] Same\: base \\[2ex] Equate\: the\: exponents \\[2ex] \dfrac{1 - 4x}{5} = x^2 \\[3ex] 1 - 4x = 5x^2 \\[3ex] 0 = 5x^2 + 4x - 1 \\[3ex] 5x^2 + 4x - 1 = 0 \\[3ex] 5x^2 + 5x - x - 1 = 0 \\[3ex] 5x(x + 1) - 1(x + 1) = 0 \\[3ex] x + 1 = 0\:\:\: OR\:\:\: 5x - 1 = 0 \\[3ex] x = -1\:\:\: OR\:\:\: 5x = 1 \\[3ex] x = -1\:\:\: OR\:\:\: x = \dfrac{1}{5} \\ $
Second Method - By Logarithms
$\left(\sqrt[5]{7}\right)$$1 - 4x$ = $7^{x^2}$

$\sqrt[5]{7}$ = $7$$\dfrac{1}{5}$ $...Law\: 7...Exp$

$\left(\sqrt[5]{7}\right)$$1 - 4x$ = $7$$\left(\dfrac{1}{5}\right) * (1 - 4x)$

$7$$\left(\dfrac{1}{5}\right) * (1 - 4x)$ = $7$$\left(\dfrac{1 - 4x}{5}\right)$ $...Law\: 7...Exp$

$ \rightarrow 7^{\left(\dfrac{1 - 4x}{5}\right)} = 7^{x^2} \\[3ex] Introduce\: \log_7\: to\: both\: sides \\[3ex] \rightarrow \log_7{7^{\left(\dfrac{1 - 4x}{5}\right)}} = \log_7{7^{x^2}} \\[3ex] \log_7{7^{\left(\dfrac{1 - 4x}{5}\right)}} = \left(\dfrac{1 - 4x}{5}\right) \log_7{7} ... Law\: 5...Log \\[3ex] \log_7{7^{x^2}} = x^2 \log_7{7} ... Law\: 5...Log \\[3ex] \log_7{7} = 1...Law\: 4...Log \\[3ex] \rightarrow \dfrac{1 - 4x}{5} = x^2 \\[3ex] 1 - 4x = 5x^2 \\[3ex] 0 = 5x^2 + 4x - 1 \\[3ex] 5x^2 + 4x - 1 = 0 \\[3ex] 5x^2 + 5x - x - 1 = 0 \\[3ex] 5x(x + 1) - 1(x + 1) = 0 \\[3ex] x + 1 = 0\:\:\: OR\:\:\: 5x - 1 = 0 \\[3ex] x = -1\:\:\: OR\:\:\: 5x = 1 \\[3ex] x = -1\:\:\: OR\:\:\: x = \dfrac{1}{5} $
Check

LHS
$\left(\sqrt[5]{7}\right)$$1 - 4x$

$x = -1$

$1 - 4x = 1 - 4(-1) = 1 + 4 = 5$

$\left(\sqrt[5]{7}\right)$$5$

$ \left(\sqrt[5]{7}\right) = 7^{\dfrac{1}{5}} ...Law\: 7...Exp \\[3ex] \left(\sqrt[5]{7}\right)^5 = 7^{\left(\dfrac{1}{5}\right)5} \\[3ex] 7^{\left(\dfrac{1}{5}\right)5} = 7^{\left(\dfrac{1}{5} * 5\right)} ...Law\: 5...Exp \\[3ex] = 7^1 \\[3ex] = 7 $


LHS
$\left(\sqrt[5]{7}\right)$$1 - 4x$

$x = \dfrac{1}{5}$

$1 - 4x = 1 - 4 * \dfrac{1}{5} = 1 - \dfrac{4}{5} = \dfrac{1}{5}$

$= \left(\sqrt[5]{7}\right)$$\dfrac{1}{5}$

$ \left(\sqrt[5]{7}\right) = 7^{\dfrac{1}{5}} ...Law\: 7...Exp \\[3ex] = \left(\sqrt[5]{7}\right)^5 = 7^{\left(\dfrac{1}{5}\right)\dfrac{1}{5}} \\[3ex] 7^{\left(\dfrac{1}{5}\right)\dfrac{1}{5}} = 7^{\left(\dfrac{1}{5} * \dfrac{1}{5}\right)} ...Law\: 5...Exp \\[3ex] = 7^{\dfrac{1}{25}} $

RHS
$ 7^{x^2} \\[3ex] x = -1 \\[3ex] x^2 = (-1)^2 = 1 \\[3ex] = 7^1 \\[3ex] = 7 $


RHS
$ 7^{x^2} \\[3ex] x = \dfrac{1}{5} \\[3ex] x^2 = \left(\dfrac{1}{5}\right)^2 \\[3ex] \left(\dfrac{1}{5}\right)^2 = \dfrac{1}{25} \\[3ex] = 7^{\dfrac{1}{25}} $

(9.) $\dfrac{\sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}}}{e^{2x} \div e^{-2x}} = e^{13}$


$ \dfrac{\sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}}}{e^{2x} \div e^{-2x}} = e^{13} \\[5ex] LCD = e^{2x} \div e^{-2x} \\[5ex] Multiply\: both\: sides\: by\: \left(e^{2x} \div e^{-2x}\right) \\[5ex] \left(e^{2x} \div e^{-2x}\right) * \dfrac{\sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}}}{e^{2x} \div e^{-2x}} = \left(e^{2x} \div e^{-2x}\right) * e^{13} \\[5ex] = \sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}} = \left(e^{2x} \div e^{-2x}\right) * e^{13} \\[5ex] \left(e^{2x} \div e^{-2x}\right) = \dfrac{e^{2x}}{e^{-2x}} \\[5ex] \dfrac{e^{2x}}{e^{-2x}} = e^{2x - (-2x)} ...Law\: 2...Exp \\[5ex] e^{2x - (-2x)} = e^{2x + 2x} = e^{4x} \\[5ex] \therefore \left(e^{2x} \div e^{-2x}\right) = e^{4x} \\[5ex] \sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}} = \left[\left(e^{2x} * e^{-5x}\right)^{-4}\right]^{\dfrac{1}{2}} ...Law\: 7...Exp \\[5ex] \left[\left(e^{2x} * e^{-5x}\right)^{-4}\right]^{\dfrac{1}{2}} = \left[\left(e^{2x} * e^{-5x}\right)^{-4 * \dfrac{1}{2}}\right] ...Law\: 5...Exp \\[5ex] \left[\left(e^{2x} * e^{-5x}\right)^{-4 * \dfrac{1}{2}}\right] = \left[\left(e^{2x} * e^{-5x}\right)^{-2}\right] \\[5ex] \left(e^{2x} * e^{-5x}\right) = e^{2x + (-5x)} ...Law\: 1...Exp \\[5ex] e^{2x + (-5x)} = e^{2x - 5x} = e^{-3x} \\[5ex] \left[\left(e^{2x} * e^{-5x}\right)^{-2}\right] = \left(e^{-3x}\right)^{-2} \\[5ex] \left(e^{-3x}\right)^{-2} = \left(e^{-3x * -2}\right) ...Law\: 5...Exp \\[5ex] \left(e^{-3x * -2}\right) = e^{6x} \\[5ex] \therefore \sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}} = e^{6x} \\[5ex] \rightarrow e^{6x} = e^{4x} * e^{13} \\[5ex] Divide\: both\: sides\: by\: e^{4x} \\[5ex] e^{6x} \div e^{4x} = e^{13} \\[5ex] e^{6x} \div e^{4x} = e^{6x - 4x} = e^{2x} ...Law\: 2...Exp \\[5ex] \rightarrow e^{2x} = e^{13} \\[5ex] Same\: base \\[3ex] Equate\: the\: exponents \\[3ex] 2x = 13 \\[3ex] x = \dfrac{13}{2} $
Check

LHS
$ \dfrac{\sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}}}{e^{2x} \div e^{-2x}} \\[5ex] x = \dfrac{13}{2} \\[5ex] = \dfrac{\sqrt{\left(e^{\left(2 * \dfrac{13}{2}\right)} * e^{\left(-5 * \dfrac{13}{2}\right)}\right)^{-4}}}{e^{\left(2 * \dfrac{13}{2}\right)} \div e^{\left(-2 * \dfrac{13}{2}\right)}} \\[9ex] = \dfrac{\sqrt{\left(e^{13} * e^{\dfrac{-65}{2}}\right)^{-4}}}{e^{13} \div e^{-13}} \\[9ex] e^{13} * e^{\dfrac{-65}{2}} = e^{\left(13 + \dfrac{-65}{2}\right)}...Law\: 1...Exp \\[5ex] e^{13 + \dfrac{-65}{2}} = e^{\left(13 - \dfrac{65}{2}\right)} \\[5ex] e^{\left(13 - \dfrac{65}{2}\right)} = e^{\left(\dfrac{26}{2} - \dfrac{65}{2}\right)} \\[5ex] e^{\left(\dfrac{26}{2} - \dfrac{65}{2}\right)} = e^{\left(\dfrac{26 - 65}{2}\right)} = e^{\dfrac{-39}{2}} \\[5ex] \left(e^{\dfrac{-39}{2}}\right)^{-4} = e^{\left(\dfrac{-39}{2} * -4\right)} ...Law\: 5...Exp \\[5ex] e^{\left(\dfrac{-39}{2} * -4\right)} = e^{78} \\[5ex] \sqrt{e^{78}} = \left(e^{78}\right)^{\dfrac{1}{2}} ...Law\: 7...Exp \\[5ex] \left(e^{78}\right)^{\dfrac{1}{2}} = \left(e^{78}\right)^{\dfrac{1}{2}} \\[5ex] \left(e^{78}\right)^{\dfrac{1}{2}} = e^{\left(78 * \dfrac{1}{2}\right)} ...Law\: 5...Exp \\[5ex] e^{\left(78 * \dfrac{1}{2}\right)} = e^{39} \\[5ex] Numerator = e^{39} \\[5ex] e^{13} \div e^{-13} = e^{\left(13 - (-13)\right)} ...Law\: 2...Exp \\[5ex] e^{\left(13 - (-13)\right)} = e^{\left(13 + 13\right)} = e^{26} \\[5ex] Denominator = e^{26} \\[5ex] = \dfrac{e^{39}}{e^{26}} \\[5ex] = e^{\left(39 - 26\right)} ...Law\: 2...Exp \\[5ex] = e^{13} $

RHS
$ e^{13} $

(10.) $4^{x - 4} = 64(3^x)$


$4^{x - 4} = 64(3^x)$

$3$ is not a multiple of $4$

In that regard, we shall not be solving it By Exponents.

We shall solve it By Logarithms.

$ Introduce\: \log_4\: to\: both\: sides \\[3ex] \log_4{4^{x - 4}} = \log_4{[64(3^x)]} \\[3ex] \log_4{4^{x - 4}} = (x - 4)\log_4{4} ...Law\: 1...Log \\[3ex] \log_4{4} = 1 ...Law\: 4...Log \\[3ex] (x - 4)\log_4{4} = (x - 4) * 1 = (x - 4) \\[3ex] \log_4{[64(3^x)]} = \log_4{64} + \log_4{3^x} \\[3ex] \log_4{64} = \log_4{4^3} = 3\log_4{4} ...Law\: 5...Log \\[3ex] \log_4{64} = 3 * 1 = 3 \\[3ex] \log_4{3^x} = x\log_4{3} \\[3ex] \rightarrow x - 4 = 3 + x\log_4{3} \\[3ex] x - x\log_4{3} = 3 + 4 \\[3ex] x(1 - \log_4{3}) = 7 \\[3ex] x = \dfrac{7}{1 - \log_4{3}} $

OR

$ \log_4{3} = \dfrac{\log4}{\log3} ...Law\: 6...Log \\[3ex] \log_4{3} = 0.792481250 \\[3ex] x = \dfrac{7}{1 - \log_4{3}} \\[3ex] x = \dfrac{7}{1 - 0.792481250} \\[3ex] x = \dfrac{7}{0.207518750} \\[3ex] x = 33.73189176 $
Check

LHS
$ 4^{x - 4} \\[3ex] x = 33.73189176 \\[3ex] 4^{33.73189176 - 4} \\[3ex] 4^{29.73189176} \\[3ex] 7.950281244 * 10^{17} $

RHS
$ 64(3^x) \\[3ex] 64(3^{33.73189176}) \\[3ex] 64 * 1.242231451 * 10^{16} \\[3ex] 7.950281287 * 10^{17} $

(11.) For the function $f(x) = 4(2)^x - 6$,
(a.) Write the domain.
(b.) Write the range.
(c.) Determine the asymptotes.
(d.) Graph the function. Indicate the asymptote on the graph.
Which is the graph of $f(x) = 4(2)^x - 6$ (and its asymptote)?
Question 11-1
Question 11-2
Question 11-3
Question 11-4


An exponential function is defined for all the real values of the independent variable
Therefore, all the real values of $x$ are defined in the domain.

Because all the real values of $x$ are in the domain of an exponential function, an exponential function does not have a vertical asymptote.

$ f(x) = 4(2)^x - 6 \\[3ex] (a.) \\[3ex] \underline{Domain} \\[3ex] D = \{x: x \in \mathbb{R}\} \\[3ex] D = (-\infty, \infty) \\[3ex] (b.) \\[3ex] f(x) = 4(2)^x - 6 \\[3ex] f(x) = 4 * 2^x - 6 \\[3ex] \underline{Positive\;\;Domain} \\[3ex] As \;\;x \rightarrow \infty, \;\; 2^x \rightarrow \infty \\[3ex] \underline{Negative\;\;Domian} \\[3ex] However:\;\; As \;\;x \rightarrow -\infty, \;\; 2^x \rightarrow 0 \\[3ex] How? \\[3ex] Some\;\;examples: \\[3ex] 2^{-5} = \dfrac{1}{2^5} = \dfrac{1}{32} = 0.03125 \\[5ex] 2^{-20} = \dfrac{1}{2^{20}} = \dfrac{1}{1048576} = 0.0000009536743164 \\[5ex] Would\;\;you\;\;like\;\;to\;\;try\;\; 2^{-150}? \\[3ex] approaches\;\;zero\;\;though\;\;still\;\;greater\;\;than\;\;zero \\[3ex] \implies \\[3ex] f(x) \;\;greater\;\;than\;\; 4 * 0 - 6 \\[3ex] f(x) \;\;greater\;\;than\;\; 0 - 6 \\[3ex] f(x) \;\;greater\;\;than\;\; -6 \\[3ex] \underline{Range} \\[3ex] R = \{y: y \gt -6\} \\[3ex] R = (-6, \infty) \\[3ex] (c.) \\[3ex] No\;\;vertical\;\;asymptote \\[3ex] \underline{Horizontal\;\;Asymptote} \\[3ex] f(x) = 4 * 2^x - 6 \\[3ex] As \;\;x \rightarrow -\infty, \;\; 2^x \rightarrow 0 \\[3ex] \implies \\[3ex] f(x) = 4 * 0 - 6 \\[3ex] f(x) = 0 - 6 \\[3ex] f(x) = -6 \\[3ex] HA:\;\; y = -6 \\[3ex] (d.) \\[3ex] f(x) = 4 * 2^x - 6 \\[3ex] Verify\;\;at\;\;least\;\;one\;\;point\;\;on\;\;the\;\;graph \\[3ex] when\;\;x = 0 \\[3ex] f(0) = 4 * 2^0 - 6 \\[3ex] f(0) = 4(1) - 6 \\[3ex] f(0) = 4 - 6 \\[3ex] f(0) = -2 \\[3ex] (0, -2) \;\;is\;\;a\;\;point\;\;on\;\;the\;\;graph \\[3ex] Also: \\[3ex] $ Because the coefficient of $2^x$ is $4$, the function is increasing
With a horizontal asymptote of $-6$, the graph is:
Number 11
(12.) $x^7 = 128$


First Method - By Exponents $ x^7 = 128 \\[2ex] x^7 = 2^7 \\[2ex] Exponents\: are\: the\: same \\[2ex] Equate\: the\: bases \\[2ex] x = 2 $
Second Method - By Logarithms $ x^7 = 128 \\[2ex] Introduce\: \log_2\: to\: both\: sides \\[2ex] \log_2{x^7} = \log_2{128} \\[2ex] \log_2{x^7} = 7\log_2{x} ...Law\: 5...Log \\[2ex] \log_2{128} = \log_2{2^7} = 7\log_2{2} ... Law\: 5...Log \\[2ex] \log_2{2} = 1...Law\: 4...Log \\[2ex] \log_2{128} = 7 * 1 = 7 \\[2ex] \rightarrow 7\log_2{x} = 7 \\[2ex] Divide\: both\: sides\: by\: 7 \\[2ex] \log_2{x} = 1 \\[2ex] 1 = \log_2{2} ...Law\: 4...Log \\[2ex] \log_2{x} = \log_2{2} \\[2ex] LogBase\: is\: the\: same \\[2ex] Equate\: the\: terms \\[2ex] x = 2 $
Check

LHS
$ x^7 \\[2ex] 2^{7} \\[2ex] 128 $

RHS
$ 128 $

(13.) State the:
(a.) Domain.
(b.) Vertical Asymptote.
(c.) End-behavior.
of the function $g(x) = \ln(5x + 10) + 1.3$
(d.) Which of the following graphs best represents the graph of $g(x)$?
Question 13-1
Question 13-2


The domain of a logarithmic function are all the values of $x$ for which the function has an output.
To find the domain of a logarithmic function, the argument of the function must be greater than zero.
To determine the vertical asymptote of a logarithmic function, set the argument to zero and solve for $x$

$ g(x) = \ln(5x + 10) + 1.3 \\[3ex] (a.) \\[3ex] \underline{Domain} \\[3ex] 5x + 10 \gt 0 \\[3ex] 5x \gt 0 - 10 \\[3ex] 5x \gt -10 \\[3ex] x \gt -\dfrac{10}{5} \\[5ex] x \gt -2 \\[3ex] D = \{x: x \gt -2\} \\[3ex] D = (-2, \infty) \\[3ex] (b.) \\[3ex] \underline{Vertical\;\;Asymptote} \\[3ex] 5x + 10 = 0 \\[3ex] 5x = 0 - 10 \\[3ex] 5x = -10 \\[3ex] x = -\dfrac{10}{5} \\[5ex] x = -2 \\[3ex] VA:\;\;x = -2 \\[3ex] (c.) \\[3ex] VA:\;\;x = -2 \\[3ex] As\;\;x \rightarrow -2,\;\; g(x) \rightarrow -\infty \\[3ex] As\;\;x \rightarrow \infty,\;\; g(x) \rightarrow \infty \\[3ex] $ (d.)
With a vertical asymptote of $-2$, the graph is:
Number 13
(14.) ACT For what value of $x$ is the equation $2^{2x + 7} = 2^{15}$ true?


$ 2^{2x + 7} = 2^{15} \\[3ex] Same\:\: base \\[3ex] Equate\:\: exponents \\[3ex] 2x + 7 = 15 \\[3ex] 2x = 15 - 7 \\[3ex] 2x = 8 \\[3ex] x = \dfrac{8}{2} \\[5ex] x = 4 \\[3ex] $ Check

$ \underline{LHS} \\[3ex] 2^{2x + 7} \\[3ex] x = 4 \\[3ex] 2^{2(4) + 7} \\[3ex] 2^{8 + 7} \\[3ex] 2^{15} $

$ \underline{RHS} \\[3ex] 2^{15} $