# Help with Logarithmic Equations

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• November 24th 2008, 02:48 PM
epetrik
Help with Logarithmic Equations
Please help me in anyway figure these out, I have a test tomorrow!

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!
• November 24th 2008, 03:09 PM
masters
Quote:

Originally Posted by epetrik
Please help me in anyway figure these out, I have a test tomorrow!

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!

Here's the way to go about this.

$8^{x+1}=6$

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

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

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

• November 24th 2008, 03:19 PM
guywhohatesmath
Quote:

Originally Posted by masters
Here's the way to go about this.

$8^{x+1}=6$

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

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

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

i have some more-or-less similar log problems. could you help me out as well? http://www.mathhelpforum.com/math-he...ogarithms.html
• November 24th 2008, 04:45 PM
epetrik
Thank you!!