SOLUTION: 1+log bas4 (X-1)=log base2 (X-9)

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Question 824222: 1+log bas4 (X-1)=log base2 (X-9)
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--

YOUR PROBLEM:
Sove for x.
1%2Blog%284%2C%28x-1%29%29=log%282%2C%28x-9%29%29

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

+1%2B%28log%282%2C%28x-1%29%29%29%2F%28log%282%2C4%29%29=%28log%282%2C%28x-9%29%29%29

Notice that log%282%2C4%29=2 because 2%5E2=4. So, we simplify,

1%2B%28log%282%2C%28x-1%29%29%29%2F2=log%282%2C%28x-9%29%29
2%2Blog%282%2C%28x-1%29%29=2%2Alog%282%2C%28x-9%29%29
2=2%2Alog%282%2C%28x-9%29%29-log%282%2C%28x-1%29%29

Apply the Log Law for Subtraction/Division and the Log Law for Powers
2=log%282%2C%28%28x-9%29%5E2%2F%28x-1%29%29%29

Translate the logarithmic equation into the related exponential equation.
%28x-9%29%5E2%2F%28x-1%29=2%5E2=4

Simplify.
x%5E2-18x%2B81=4%2A%28x-1%29
x%5E2-18x%2B81=4x-4
x%5E2-22x%2B85=0}

Solve this quadratic equation by factoring.
%28x-17%29%28x-5%29=0
x=17 OR x=5

Always check your answers for extraneous roots.

For x = 17:

1%2Blog%284%2C%28%2817%29-1%29%29=log%282%2C%28%2817%29-9%29%29
1%2Blog%284%2C16%29=log%282%2C8%29
1%2B2=3
3=3 TRUE

For x = 5:

1%2Blog%284%2C%28%285%29-1%29%29=log%282%2C%28%285%29-9%29%29
1%2Blog%284%2C4%29=log%282%2C-4%29

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