Exponential Functions

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College Algebra › Exponential Functions

Questions 1 - 9

1

Solve the following:

Explanation

To solve the following, you must "undo" the 5 with taking log based 5 of both sides. Thus,

$\log_5{5^x}=\log_5{125}$

$x=\log_5{125}$

The right hand side can be simplified further, as 125 is a power of 5. Thus,

$x=\log_5{125}$

$x=\log_5{5^3}$

2

Which equation is equivalent to:

$\log _{4}\left( \frac{1}{2}\right)=x$

$4^{x}=\frac{1}{2}$

$\frac{1^{4}}{2}=x^{4}$

$\left(\frac{1}{2}\right)^{4}=x$

$4^{\frac{1}{2}}=x$

$(4)^{x}=\left(\frac{1}{2}\right)^{x}$

Explanation

$log{_{a}}^{b}=x$ ,

$\Rightarrow a^{x}=b$

So, ${{log_{4}}^{\frac{1}{2}}}=x\Rightarrow 4^{x}=\frac{1}{2}$

3

Rewrite the following expression as an exponential expression:

$3^{14,000}=b$

$b^{14,000}=3$

Explanation

Rewrite the following expression as an exponential expression:

Recall the following property of logs and exponents:

Can be rewritten in the following form:

So, taking the log we are given;

We can rewrite it in the form:

$3^{14,000}=b$

So b must be a really huge number!

4

Solve:

$\frac{1}{2x+3}$

The answer does not exist.

Explanation

To solve , it is necessary to know the property of .

Since and the terms cancel due to inverse operations, the answer is what's left of the term.

The answer is:

5

Solve for :

$2^{x}+ 2 ^{x+2} = 171$

(Nearest hundredth)

$x \approx 5.10$

$x \approx 2.71$

$x \approx 3.65$

The equation has no solution.

$x \approx 2.84$

Explanation

Apply the Product of Powers Property to rewrite the second expression:

$2^{x}+ 2 ^{x+2} = 171$

$2^{x}+ 2 ^{2} \cdot 2 ^{x} = 171$

$1 \cdot 2^{x}+ 4 \cdot 2 ^{x} = 171$

Distribute out:

$(1 + 4) \cdot 2 ^{x} = 171$

$5 \cdot 2 ^{x} = 171$

Divide both sides by 5:

$\frac{5 \cdot 2 ^{x}}{5} = \frac{171}{5}$

$2 ^{x} = 34.2$

Take the natural logarithm of both sides (and note that you can use common logarithms as well):

$\ln 2 ^{x} = \ln 34.2$

Apply a property of logarithms:

$x \ln 2 = \ln 34.2$

Divide by $\ln 2$ and evaluate:

$\frac{x \ln 2}{\ln2} = \frac{\ln 34.2}{\ln2}$

$x = \frac{3.5322 }{0.6931 }$

$x \approx 5.10$

6

Solve for :

$3^{2x+ 1} = 9^{2x- 4 }$

(Nearest hundredth, if applicable).

The equation has no solution.

Explanation

$9 = 3^{2}$ , so rewrite the expression at right as a power of 3 using the Power of a Power Property:

$3^{2x+ 1} = 9^{2x- 4 }$

$3^{2x+ 1} = (3^{2} )^{2x- 4 }$

$3^{2x+ 1} = 3^{2(2x- 4 )}$

Set the exponents equal to each other and solve the resulting linear equation:

Distribute:

Subtract and 1 from both sides; we can do this simultaneously:

Divide by :

$\frac{- 2x}{-2} = \frac{-9}{-2}$

7

Convert the following logarithmic equation to an exponential equation.

$3^{156}=14x$

$3^{14x}=156$

$(14x)^{156}=3$

Explanation

Convert the following logarithmic equation to an exponential equation.

To convert from logarithms to exponents, recall the following property:

Can be rewritten as:

So, starting with

,

We can get

$3^{156}=14x$

8

Convert the following logarithmic equation to an exponential equation:

$log_{13}b=147$

$13^{147}=b$

$13^{b}=147$

$b^{13}=147$

Explanation

Convert the following logarithmic equation to an exponential equation:

$log_{13}b=147$

Recall the following:

This

Can be rewritten as

So, our given logarithm

$log_{13}b=147$

Can be rewritten as

$13^{147}=b$

Fortunately we don't need to expand, because this woud be a very large number!

9

$Log_{b}x=y$

What is the inverse of the log function?

$b^{y}=x$

$b^{x}=y$

$x^{b}=y$

$y^{b}=x$

$b=x\cdot y$

Explanation

This is a general formula that you should memorize. The inverse of $Log_{b}x=y$ is $b^{y}=x$ . You can use this formula to change an equation from a log function to an exponential function.

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