SOLUTION: Hi!
Please help! On my worksheet it says to solve this question:
logBasec(x-1)-logBasea(x+6)=logBasea(x-2)-logBasea(x+3)
I'm not sure how to do this - also, how do I 'solve' i
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Hi!
Please help! On my worksheet it says to solve this question:
logBasec(x-1)-logBasea(x+6)=logBasea(x-2)-logBasea(x+3)
I'm not sure how to do this - also, how do I 'solve' i
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Question 864871: Hi!
Please help! On my worksheet it says to solve this question:
logBasec(x-1)-logBasea(x+6)=logBasea(x-2)-logBasea(x+3)
I'm not sure how to do this - also, how do I 'solve' if there are two variables? Could someone show me how to do this step-by-step? Thank you so much! Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! logBasec(x-1)-logBasea(x+6)=logBasea(x-2)-logBasea(x+3)
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loga(x-1) - loga(x+6) = loga(x-2)-loga(x+3)
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(x-1)/(x+6) = (x-2)/(x+3)
Cross-multiply to get:
(x-1)(x+3) = (x+6)(x-2)
x^2 + 2x -3 = x^2 + 4x - 12
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2x-3 = 4x-12
2x = 9
x = 9/2
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Cheers,
Stan H.
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