SOLUTION: Please help me solve: log base 5 (7+x) + log base 5 (8-x) - log base 5 2 = 2
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Question 103958: Please help me solve: log base 5 (7+x) + log base 5 (8-x) - log base 5 2 = 2
Found 2 solutions by stanbon, scott8148:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
log base 5 (7+x) + log base 5 (8-x) - log base 5 2 = 2
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Keep the base 5 in mind.
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log[(7+x)(8-x)/2] = 2
log[56+x-x^2] = 2
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Now, using the fact that the base is 5 you get:
56+x-x^2 = 5^2
x^2-x-31=0
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x = [1 +- sqrt(1-4*-31]/2
x = [1 +- sqrt(125)]/2
x = [1 +- 5sqrt(5)]/2
x = [1 + 5sqrt(5)]/2 = 6.09...
OR
x = [1 - 5sqrt(5)]/2 = -5.09...
Check these answers in the original equation
to see if either is extraneous.
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Cheers,
Stan H.
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
(7+x)(8-x)/2=25 ... 56+x-x^2=50 ... x^2-x-6=0 ... (x-3)(x+2)=0 ... x=3 and x=-2
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