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Algebra II : Log-Base-10

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Example Questions

Example Question #41 : Logarithms

Solve: $\displaystyle 2log(300)$

Possible Answers:

$\displaystyle 4log(2)+2$

$\displaystyle 2log(3)+2$

$\displaystyle 3log(2)+2$

$\displaystyle 2log(3)+4$

$\displaystyle 3log(2)+4$

Correct answer:

$\displaystyle 2log(3)+4$

Explanation:

Break up $\displaystyle log(300)$ using log rules. The log has a default base of ten.

$log(A\times B) = log(A)+log(B)$ $\displaystyle log(A\times B) = log(A)+log(B)$

$log(300) = log(3)+log(100)=log_{10}(3)+log_{10}(10^2)$ $\displaystyle log(300) = log(3)+log(100)=log_{10}(3)+log_{10}(10^2)$

The exponent can be brought down as the coefficient since the bases of the second term are common.

$log_{10}(3)+log_{10}(10^2) =log_{10}(3)+2log_{10}(10) = log(3)+2$ $\displaystyle log_{10}(3)+log_{10}(10^2) =log_{10}(3)+2log_{10}(10) = log(3)+2$

This means that: $\displaystyle 2log(300)=2[ log(3)+2] = 2log(3)+4$

The answer is: $\displaystyle 2log(3)+4$

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Example Question #22 : Log Base 10

Evaluate: $2log_{10}(\frac{1}{1000})$ $\displaystyle 2log_{10}(\frac{1}{1000})$

Possible Answers:

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

$\displaystyle -3$

$\frac{1}{5}$ $\displaystyle \frac{1}{5}$

$\frac{1}{50}$ $\displaystyle \frac{1}{50}$

$\displaystyle -6$

Correct answer:

$\displaystyle -6$

Explanation:

We will need to write fraction in terms of the given base of log, which is ten.

$2log_{10}(\frac{1}{1000})=2log_{10}(10^{-3})$ $\displaystyle 2log_{10}(\frac{1}{1000})=2log_{10}(10^{-3})$

According to the log rules:

$log_{x}(x^Y) =Y$ $\displaystyle log_{x}(x^Y) =Y$

This means that the expression of log based 10 and the power can be simplified.

$2log_{10}(10^{-3}) =2\cdot -3 = -6$ $\displaystyle 2log_{10}(10^{-3}) =2\cdot -3 = -6$

The answer is: $\displaystyle -6$

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Example Question #281 : Mathematical Relationships And Basic Graphs

Evaluate $\log_{8}172$ $\displaystyle \log_{8}172$ to the nearest tenth.

Possible Answers:

2.1 $\displaystyle 2.1$

2.9 $\displaystyle 2.9$

2.3 $\displaystyle 2.3$

2.5 $\displaystyle 2.5$

2.7 $\displaystyle 2.7$

Correct answer:

2.5 $\displaystyle 2.5$

Explanation:

Since most calculators only have common and natural logarithm keys, this can best be solved as follows:

By the Change of Base Property of Logarithms, if $\displaystyle M,N,a > 0$ and $a, M \ne 1$ $\displaystyle a, M \ne 1$,

$\log_{M}N = \frac{\log_{a}N}{\log_{a}M}$ $\displaystyle \log_{M}N = \frac{\log_{a}N}{\log_{a}M}$

Setting $\displaystyle M = 8, N = 172, a = 10$, we can restate this logarithm as the quotient of two common logarithms, and calculate accordingly:

$\log_{8}172 = \frac{\log 172}{\log 8}\approx \frac{2.2355 }{0.9031}\approx 2.4754$ $\displaystyle \log_{8}172 = \frac{\log 172}{\log 8}\approx \frac{2.2355 }{0.9031}\approx 2.4754$

or, when rounded, 2.5.

This can also be done with natural logarithms, yielding the same result.

$\log_{8}172 = \frac{\ln 172}{\ln 8} \approx \frac{5.1475}{2.0794}\approx 2.4754$ $\displaystyle \log_{8}172 = \frac{\ln 172}{\ln 8} \approx \frac{5.1475}{2.0794}\approx 2.4754$

Rond to one decimal place.

2.5 $\displaystyle 2.5$

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Example Question #41 : Understanding Logarithms

Solve for $\displaystyle x$ in the equation:

$\displaystyle -log(x)=4.2$

Possible Answers:

$x=e^{-4.2}$ $\displaystyle x=e^{-4.2}$

$x=10^{-4.2}$ $\displaystyle x=10^{-4.2}$

$x=10^{-log(4.2)}$ $\displaystyle x=10^{-log(4.2)}$

$x=e^{4.2}$ $\displaystyle x=e^{4.2}$

Correct answer:

$x=10^{-4.2}$ $\displaystyle x=10^{-4.2}$

Explanation:

This question tests your understanding of log functions.

$\displaystyle log(x)=y$ can be converted to the form $x=10^{y}$ $\displaystyle x=10^{y}$.

In this problem, make sure to divide both sides by $\displaystyle -1$ in order to put it in the above form, where $\displaystyle log(x)=-4.2$. Remember $log(x)=log_{10}(x)$ $\displaystyle log(x)=log_{10}(x)$.

Therefore,

$log_{10}(x)=-4.2 \rightarrow x=10^{-4.2}$ $\displaystyle log_{10}(x)=-4.2 \rightarrow x=10^{-4.2}$

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Example Question #281 : Mathematical Relationships And Basic Graphs

Evaluate $\displaystyle x$.

$5 ^{x} = 12$ $\displaystyle 5 ^{x} = 12$

Possible Answers:

$x= \frac{ \log 5}{12}$ $\displaystyle x= \frac{ \log 5}{12}$

$x= \frac{ \log 12 }{5}$ $\displaystyle x= \frac{ \log 12 }{5}$

$x= \log 7$ $\displaystyle x= \log 7$

$x=\frac{ \log 5}{\log 12}$ $\displaystyle x=\frac{ \log 5}{\log 12}$

$x=\frac{\log 12}{\log 5}$ $\displaystyle x=\frac{\log 12}{\log 5}$

Correct answer:

$x=\frac{\log 12}{\log 5}$ $\displaystyle x=\frac{\log 12}{\log 5}$

Explanation:

Take the common logarithm of both sides, and take advantage of the property of the logarithm of a power:

$5 ^{x} = 12$ $\displaystyle 5 ^{x} = 12$

$\log 5 ^{x} = \log 12$ $\displaystyle \log 5 ^{x} = \log 12$

$x \log 5 = \log 12$ $\displaystyle x \log 5 = \log 12$

$\frac{x \log 5 }{\log 5 }=\frac{ \log 12}{\log 5}$ $\displaystyle \frac{x \log 5 }{\log 5 }=\frac{ \log 12}{\log 5}$

$x=\frac{ \log 12}{\log 5}$ $\displaystyle x=\frac{ \log 12}{\log 5}$

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Algebra II : Log-Base-10

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #41 : Logarithms

Example Question #22 : Log Base 10

Example Question #281 : Mathematical Relationships And Basic Graphs

Example Question #41 : Understanding Logarithms

Example Question #281 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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