# Math Help - Help with Logarithmic Equations

1. ## Help with Logarithmic Equations

Solve the equation:
8^x+1=6

So far I have:
(x+1)log8 = log6
= xlog8 + 1log8 = log6

Do I subtract the 1log8 from both sides next?
Really confused so any help would be greatly appreciated!

2. Originally Posted by epetrik

Solve the equation:
8^x+1=6

So far I have:
(x+1)log8 = log6
= xlog8 + 1log8 = log6

Do I subtract the 1log8 from both sides next?
Really confused so any help would be greatly appreciated!

$8^{x+1}=6$

$(x+1)\log 8=\log 6$

$x+1=\frac{\log 6}{\log 8}$

$x=\frac{\log 6}{\log 8}-1$

3. Originally Posted by masters
$8^{x+1}=6$
$(x+1)\log 8=\log 6$
$x+1=\frac{\log 6}{\log 8}$
$x=\frac{\log 6}{\log 8}-1$