Most of us don't know what the exponent would be if \(\displaystyle 7^x=118\) and unfortunately there is no \(\displaystyle \log_7\) on a graphing calculator -- only \(\displaystyle \log\) (which stands for \(\displaystyle \log_{10}\)).

Fortunately we can use the base change rule: \(\displaystyle \log_ba=\frac{\log_{10}a}{\log_{10}b}\)

Plug in our given values.

\(\displaystyle \log_7118=\frac{\log_{10}118}{\log_{10}7}\)

\(\displaystyle \log_7118=\frac{2.07}{0.845}\)

\(\displaystyle \log_7118=2.45\)

For any equation \(\displaystyle \log_{10}(y) = x\), \(\displaystyle 10^{x } = y\). Thus, we are trying to determine what power of 10 is 1000. \(\displaystyle 1000 = 10^3\), so our answer is 3. 

\(\displaystyle \log_xa=b\) is equivalent to \(\displaystyle x^b=a\). In this case, you know the value of \(\displaystyle a\) (the argument of a logarithmic equation) and b (the answer to the logarithmic equation). You must find a solution for the base.

\(\displaystyle x^{7}=128\)

\(\displaystyle \sqrt[7]{x}=\sqrt[7]{128}\)

\(\displaystyle x=2\)

Using properties of logs we get:

\(\displaystyle log_{5}75 - log_{5}3 = log_{5}\frac{75}{3}= log_{5}25\)

\(\displaystyle log_{5}25= log_{5}5^{2}= 2log_{5}5= 2\cdot 1=2\)

Recall the log rule: 

\(\displaystyle log(a) + log(b) = log(ab)\)

In this particular case, \(\displaystyle a = 4\) and \(\displaystyle b = 12\). Thus, our answer is \(\displaystyle log(4\cdot 12) = log(48)\).

Since the bases of the logs are the same and the logarithms are added, the arguments can be multiplied together. We then simplify the right side of the equation:

\(\displaystyle log_{6}(n-3)+log_{6}(n+2)=log_{3}(3)\)

\(\displaystyle log_{6}(n-3)(n+2)=1\)

The logarithm can be converted to exponential form:

\(\displaystyle (n-3)(n+2)=6^{1}\)

\(\displaystyle n^2-n-6=6\)

\(\displaystyle n^2-n-12=0\)

Factor the equation:

\(\displaystyle (n-4)(n+3)=0\)

\(\displaystyle n=4\)

Although there are two solutions to the equation, logarithms cannot be negative. Therefore, the only real solution is \(\displaystyle 4\).

The rule for the addition of logarithms is as follows: 

\(\displaystyle log(ab) = log(a)+log(b)\). 

As an application of this,\(\displaystyle log(5x) = log(5)+log(x)\). 

The logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator.

\(\displaystyle log_A(\frac{B}{C})=log_AB-log_AC\)

If we encounter two logarithms with the same base, we can likely combine them. In this case, we can use the reverse of the above identity.

\(\displaystyle log_AB-log_AC=log_A(\frac{B}{C})\)

\(\displaystyle A=4,\ B=7,\ C=5\)

\(\displaystyle log_47-log_45=log_4(\frac{7}{5})\)

Using the logarithm rules, exponents within logarithms can be removed and simply multiplied by the remaining logarithm.  This expression can be simplified as \(\displaystyle 5\log_2(4)\log_2(4)\)

\(\displaystyle log 2^x = 6\)

Use the power reducing theorem:

\(\displaystyle \log a^x = b \rightarrow x\log a = b\)

\(\displaystyle a=2\) and \(\displaystyle b=6\)

\(\displaystyle x\log 2 = 6\)

\(\displaystyle x= \frac{6}{\log 2}\)

\(\displaystyle x=19.93\)