Understanding Logarithms

Help Questions

Algebra II › Understanding Logarithms

Questions 1 - 10

1

Solve the following:

$\log10000+\log10$

$\log10010$

Explanation

When the base isn't explicitly defined, the log is base 10. For our problem, the first term

$\log 10000=x$

is asking:

For the second term,

$\log(10)=y$

is asking:

So, our final answer is

2

Solve: $2log_{10}(100^3)$

$\textup{Cannot be determined.}$

Explanation

Change the base of the inner term or log to base ten.

$2log_{10}(100^3)= 2log_{10}(10^{2(3)})= 2log_{10}(10^6)$

According to the log property:

$log_{b}(b^x) = x$

The log based ten and the ten to the power of will cancel, leaving just the power.

$2log_{10}(10^6) = 2(6) =12$

The answer is:

3

Solve: $2log_{10}(100^3)$

$\textup{Cannot be determined.}$

Explanation

Change the base of the inner term or log to base ten.

$2log_{10}(100^3)= 2log_{10}(10^{2(3)})= 2log_{10}(10^6)$

According to the log property:

$log_{b}(b^x) = x$

The log based ten and the ten to the power of will cancel, leaving just the power.

$2log_{10}(10^6) = 2(6) =12$

The answer is:

4

Solve the following:

$\log10000+\log10$

$\log10010$

Explanation

When the base isn't explicitly defined, the log is base 10. For our problem, the first term

$\log 10000=x$

is asking:

For the second term,

$\log(10)=y$

is asking:

So, our final answer is

5

Determine the value of: $6ln(e^{-3x})$

$-\frac{1}{2}x$

Explanation

The natural log has a default base of .

According to the rule of logs, we can use:

$log_{b}(b^x) = x$

$6ln(e^{-3x}) = 6ln_e(e^{-3x})$

The coefficient in front of the natural log can be transferred as the power of the exponent.

$6ln_e(e^{-3x}) = ln_e(e^{(-3x)(6)})$

The natural log and base e will cancel, leaving just the exponent.

The answer is:

6

Given the following:

$\log_{8} M = N$

Decide if the following expression is true or false:

$\log_{M} 8 = \frac{1}{N}$ for all positive .

True

False

Explanation

By definition of a logarithm,

$\log_{8} M = N$

if and only if

$8^{N } = M$

Take the th root of both sides, or, equivalently, raise both sides to the power of $\frac{1}{N}$ , and apply the Power of a Power Property:

$\left (8^{N } \right ) ^{\frac{1}{N}} = M ^{\frac{1}{N}}$

$8^{N \cdot \frac{1}{N} } = M ^{\frac{1}{N}}$

$8^{1 } = M ^{\frac{1}{N}}$

or

$M ^{\frac{1}{N}} = 8$

By definition, it follows that $\log_{M} 8 = \frac{1}{N}$ , so the statement is true.

7

$\log_{N }10 = M$ , with positive and not equal to 1.

Which of the following is true of $\log_{N }100$ for all such ?

$\log_{N }100 = 2M$

$\log_{N }100 = M^{2}$

$\log_{N }100 = M + 2$

$\log_{N }100 = M + 1$

$\log_{N }100 = \frac{M}{2}$

Explanation

By definition,

$\log_{N }10 = M$

If and only if

$N^{M}= 10$

Square both sides, and apply the Power of a Power Property to the left expression:

$(N^{M})^{2}= 10^{2}$

$N^{M\cdot 2}= 100$

$N^{2M}= 100$

It follows that for all positive not equal to 1,

$\log_{N }100 = 2M$

for all .

8

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

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 and ,

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

Setting , we can restate this logarithm as the quotient of two common logarithms, and calculate accordingly:

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

or, when rounded, 2.5.

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

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

Rond to one decimal place.

9

Evaluate .

Explanation

The first thing we can do is bring the exponent out of the log, to the front:

Next, we evaluate :

Recall that log without a specified base is base 10 thus

$\log_ba=c \Rightarrow b^c=a$

$\log_{10}100=c \Rightarrow 10^c=100 \Rightarrow c=2$ .

Therefore

becomes,

$3\cdot2$ .

Finally, we do the simple multiplication:

10

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

$\frac{2}{3}$

$\frac{1}{5}$

$\frac{1}{50}$

Explanation

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

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

According to the log rules:

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

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

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

The answer is:

Page 1 of 14

Return to subject