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:

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

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

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

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

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

(a^{x})^{y}=a^{xy}

\sqrt[b]{a}=a^{\frac{1}{b}}
Now I will move onto logarithmic properties

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

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

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

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

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

\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

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

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

another thing you should know is

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

a^{\log_a(x)}=x

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

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

first combien the logs to get

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

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

so we have 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 a=e^{4}

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

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

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

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

\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