Question

Please help solve this: $3^{x + 1} = 100$

Answer 1

remember, you can do anything to an equaiton as long as you do it to both sides

also
$log(a^x)=xlog(a)$
and$ln(x)=log_e(x)$
ok
if no base is stated, assume log=$log_{10}$
I'm going to use ln instead of log because ln is easier to find on a calculator

$3^{x+1}=100$
take ln of both sides
$ln(3^{x+1})=ln(100)$
$(x+1)ln(3)=ln(100)$
$xln(3)+1ln(3)=ln(100)$
minus ln(3) both sides
$xln(3)=ln(100)-ln(3)$
divide both sides by ln(3)
$x=\frac{ln(100)}{ln(3)}-1$

or if you wanted to use log base 10
$3^{x+1}=100$
take log both sides
$log(3^{x+1})=log(100)$
$log(3^{x+1})=2$
$(x+1)log(3)=2$
$xlog(3)+log(3)=2$
minus log(3) both sides
$xlog(3)=2-log(3)$
divide both sides by log(3)
$x=\frac{2}{log(3)}-1$



so
$x=\frac{ln(100)}{ln(3)}-1$
or if you wanted log
$x=\frac{2}{log(3)}-1$