# Help with Logarithmic Equations

• Nov 24th 2008, 03:48 PM
epetrik
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!
• Nov 24th 2008, 04:09 PM
masters
Quote:

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!

$8^{x+1}=6$

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

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

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

• Nov 24th 2008, 04:19 PM
guywhohatesmath
Quote:

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