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Algebra II : Solving Logarithms

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Example Questions

Example Question #11 : Solving Logarithms

Evaluate log_8128 .

Possible Answers:

$\frac{7}{3}$

$\frac{5}{2}$

No solution

Correct answer:

$\frac{7}{3}$

Explanation:

In logarithmic expressions, a^c=b is the same thing as log_ab=c .

Therefore, the equation can be rewritten as 8^x=128 .

Both 8 and 128 are powers of 2, so the equation can then be rewritten as (2^3)^n=(2^7) .

Since both sides have the same base, set 3n=7 .

Solve by dividing both sides of the equation by 3: $n=\frac{7}{3}$ .

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Example Question #12 : Solving Logarithms

Solve the equation for .

$\log_9(3-x) = \log_9(5x-15)$

Possible Answers:

No solution

x=3

x=2

$x=\frac{18}{4}$

x=-3

Correct answer:

x=3

Explanation:

Because both sides have the same logarithmic base, both terms can be set equal to each other:

3-x=5x-15

Now, evaluate the equation.

First, add x to both sides: 3=6x-15

Add 15 to both sides: 18=6x

Finally, divide by 6: x=3 .

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Example Question #11 : Solving Logarithms

Solve this logarithmic equation: $\log_{10}(5x-2)=\log_{10}(x+6)$

Possible Answers:

x=1

x=.5

x=2

None of the other answers.

x=4

Correct answer:

x=2

Explanation:

To solve this problem you must be familiar with the one-to-one logarithmic property.

$\log_{10}x=\log_{10}y$ if and only if x=y. This allows us to eliminate to logarithmic functions assuming they have the same base.

$\log_{10}(5x-2)=\log_{10}(x+6)$

one-to-one property:

(5x-2)=(x+6)

isolate x's to one side:

4x-2=6

move constant:

4x=8

x=2

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Example Question #2 : Solving Logarithmic Functions

Solve the equation:

$5e^{3x+4}-10=0$

Possible Answers:

$x=\frac{-4+ln2}{3}$

$x=-\frac{4}{3}+ln10$

No solution exists

x=-4+ln2

$x=\frac{ln6}{3}$

Correct answer:

$x=\frac{-4+ln2}{3}$

Explanation:

Get all the terms with e on one side of the equation and constants on the other.

$5e^{3x+4}-10=0$

$e^{3x+4}=2$

Apply the logarithmic function to both sides of the equation.

$ln(e^{3x+4})=ln2$

3x+4=ln2

3x=-4+ln2

$x=\frac{-4+ln2}{3}$

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Example Question #12 : Solving Logarithms

Solve the equation:

2log(x)+log(5)=3

Possible Answers:

$x=\pm 10\sqrt{2}$

$x=\pm \sqrt{995}$

$x=\pm6\sqrt{2}$

$x=\pm 12$

$x=\pm3\sqrt{5}$

Correct answer:

$x=\pm 10\sqrt{2}$

Explanation:

Recall the rules of logs to solve this problem.

First, when there is a coefficient in front of log, this is the same as log with the inside term raised to the outside coefficient.

2log(x)+log(5)=3

$log(x^{2})+log(5)=3$

Also, when logs of the same base are added together, that is the same as the two inside terms multiplied together.

In mathematical terms:

$log(a)+log(b)=log(a\cdot b)$

Thus our equation becomes,

$log(5x^{2})=3$

To simplify further use the rule,

$log_b(x)\rightarrow b^{log_b(x)}=x$

$10^{log(5x^{2})}=10^{3}$

$5x^{2}=1000$

$x^{2}=200$

$x=\pm10\sqrt{2}$ .

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Example Question #16 : Solving Logarithms

Solve for $\small x$ :

$\small \log_{2}(x+6)=4$

Possible Answers:

$\textup{No solution}$

$\frac{1}{4}$

Correct answer:

Explanation:

Logarithms are another way of writing exponents. In the general case, $\small \log_a{x}=b$ really just means $\small a^{b}=x$ . We take the base of the logarithm (in our case, 2), raise it to whatever is on the other side of the equal sign (in our case, 4) and set that equal to what is inside the parentheses of the logarithm (in our case, x+6). Translating, we convert our original logarithm equation into $\small 2^{4}=x+6$ . The left side of the equation yields 16, thus $\small x=10$ .

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Example Question #13 : Solving Logarithms

$\log_{5}x=3$

Possible Answers:

125

$\frac{1}{25}$

$\frac{1}{125}$

Correct answer:

125

Explanation:

To solve this equation, remember log rules

$\log_{b}x=y; b^{y}=x$ .

This rule can be applied here so that

5^3=x

and

125=x.

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Example Question #11 : Solving And Graphing Logarithms

$\log_{4}64-\log_{3}81=$

Possible Answers:

Correct answer:

Explanation:

To solve this equation, you must first simplify the log expressions and remember log laws ( $\log_{b}x=y$ ; $b^{y}=x$ ). Therefore, for the first expression, $4^{x}=64$ so x=3 . For the second expression, 3^x=81 so x=4. Then, substitute those numbers in so that you get 3-4=-1 .

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Example Question #19 : Solving Logarithms

Solve for

$log_{5}(x) = 2$

Possible Answers:

x = 12

x = 25

x = 32

x = 15

x = 5

Correct answer:

x = 25

Explanation:

$log_{5}(x) = 2$

Remember that a logarithm is nothing more than an exponent. The equation reads, the logarithm (exponent) with base that gives is .

x = 5^2

x = 25

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Example Question #20 : Solving Logarithms

Solve the equation

$\log_4(4x+7)=\log_4(11x)$

Possible Answers:

None of the other answers

x=2

x=3

x=0.5

x=1

Correct answer:

x=1

Explanation:

Since both sides are a log with the same base, their arguments must be equal.

Setting them equal:

4x+7=11x

Solve for x:

7=7x

1=x

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Algebra II : Solving Logarithms

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #11 : Solving Logarithms

Example Question #12 : Solving Logarithms

Example Question #11 : Solving Logarithms

Example Question #2 : Solving Logarithmic Functions

Example Question #12 : Solving Logarithms

Example Question #16 : Solving Logarithms

Example Question #13 : Solving Logarithms

Example Question #11 : Solving And Graphing Logarithms

Example Question #19 : Solving Logarithms

Example Question #20 : Solving Logarithms

All Algebra II Resources

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