Question 824222
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Hi, there--

YOUR PROBLEM:
Sove for x.
{{{1+log(4,(x-1))=log(2,(x-9))}}}

SOLUTION:
Use the change of base formula to convert the lot expressions to the same base.

{{{ 1+(log(2,(x-1)))/(log(2,4))=(log(2,(x-9)))}}}

Notice that {{{log(2,4)=2}}} because {{{2^2=4}}}. So, we simplify,

{{{1+(log(2,(x-1)))/2=log(2,(x-9))}}}
{{{2+log(2,(x-1))=2*log(2,(x-9))}}}
{{{2=2*log(2,(x-9))-log(2,(x-1))}}}

Apply the Log Law for Subtraction/Division and the Log Law for Powers
{{{2=log(2,((x-9)^2/(x-1)))}}}

Translate the logarithmic equation into the related exponential equation.
{{{(x-9)^2/(x-1)=2^2=4}}}

Simplify.
{{{x^2-18x+81=4*(x-1)}}}
{{{x^2-18x+81=4x-4}}}
{{{x^2-22x+85=0}}}}

Solve this quadratic equation by factoring.
{{{(x-17)(x-5)=0}}}
{{{x=17}}} OR {{{x=5}}}

Always check your answers for extraneous roots.

For x = 17:

{{{1+log(4,((17)-1))=log(2,((17)-9))}}}
{{{1+log(4,16)=log(2,8)}}}
{{{1+2=3}}}
{{{3=3}}} TRUE

For x = 5:

{{{1+log(4,((5)-1))=log(2,((5)-9))}}}
{{{1+log(4,4)=log(2,-4)}}}

Here we have a problem because the logarithm of a negative number is not part of the set of 
real numbers. (There is no real exponent to which you can raise the base 2 and get -4.) Thus 
x=5 is not a solution to this equation.

Hope this helps, and I'm very sorry for the delay!

Ms. Figge
math.in.the.vortex@gmial.com
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