I see all the time kids asking about the same questions so I will create this as to reference when basic methodology is needed to be taught

To start off basic exponent rules:

$\displaystyle (ab)^{x}=a^{x}\cdot{b^{x}}$

$\displaystyle a^{x}\cdot{a^{y}}=a^{x+y}$...NOTE$\displaystyle a^{x}\cdot{b^{y}}\ne{(ab)^{x+y}}$

$\displaystyle \frac{1}{a^{x}}=a^{-x}$

Using this in conjunction with the second rule we can see that

$\displaystyle \frac{a^{x}}{a^{y}}=a^{x}\cdot{a^{-y}}=a^{x+-y}=a^{x-y}$

$\displaystyle (a^{x})^{y}=a^{xy}$

$\displaystyle \sqrt[b]{a}=a^{\frac{1}{b}}$

Now I will move onto logarithmic properties

NOTE:$\displaystyle log_e(x)=\ln(x)$ is used in lieu of other logarithims for its shortness....these rules apply to all logarithims...and for knowledge

$\displaystyle lg(x)=log_2(x)$ and $\displaystyle log(x)=log_{10}(x)$

$\displaystyle \ln(ab)=\ln(a)+\ln(b)$

$\displaystyle \ln\bigg(\frac{a}{b}\bigg)=\ln(a)-\ln(b)$

$\displaystyle \ln(a^{b})=b\ln(a)$

$\displaystyle \ln(\sqrt[c]{a})=\ln(a^{\frac{1}{c}})=\frac{\ln(a)}{c}$

A very important one for solving equations as well as using a calculator is the change of base theorem which states

$\displaystyle log_a(b)=\frac{log_c(b)}{log_c(a)}$

Most useful example is $\displaystyle log_a(b)=\frac{\ln(b)}{\ln(a)}$

another thing you should know is

$\displaystyle ln_a(b)=\frac{1}{ln_b(a)}$

$\displaystyle a^{\log_a(x)}=x$

$\displaystyle log_a(a^{x})x$

Now I will go over the way to solve two basic logarithmic equations

Case 1...this is one with a pre-existing log

$\displaystyle \ln(a^3)-\ln(a)=8$

first combien the logs to get

$\displaystyle \ln\bigg(\frac{a^3}{a}\bigg)=\ln(a^2)=8$

we want to isolate the x so we introduce the exponential function($\displaystyle e^{x}$) to eliminate the log

so we have $\displaystyle e^{\ln(a^2)}\Rightarrow{a^2=e^8}$

taking the squareroot of both sides and using rules 5 and 6 in the exponent section we see the answer is $\displaystyle a=e^{4}$

Case 2....introducing logs to eliminate exponents...

$\displaystyle 8^{x}-25=0$

First thing we want to do is introduce a log to elminate the exponent...the log we introduce is always of same base as the exponent in this case 8..so we have

$\displaystyle log_8(8^x)=x=\log_8(25)$

at this point I think it is prudent to discuss extraneous solutions

$\displaystyle \log_a(x)$ is defined only for positive values for x...so you might get an algaebraic solution that works but it makes a log undefined..these solutiong are to be discarded

any questions just ask