SOLUTION: Solve. log(base4)(x+4)-log(base16)x=1

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Question 644599: Solve. log(base4)(x+4)-log(base16)x=1
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve. log(base4)(x+4)-log(base16)x=1
-----------------
log4(x+4) - log(x)/log(16) = 1
-----
log4(x+4) - log4(x)/log4(16) = 1
----
log4(x+4) - (1/4)log4(x) = 1
----
(3/4)log4(x) = 1
log4(x) = 4/3
---
x = 4^(4/3)
----
x = 6.35
===========================
Cheers,
Stan H.
===========================

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Solve. log(base4)(x+4)-log(base16)x=1


log+%284%2C+%28x+%2B+4%29%29+-+log+%2816%2C+%28x%29%29+=+1
log+%284%2C+%28x+%2B+4%29%29+-+log+%284%2C+%28x%29%29%2Flog+%284%2C+%2816%29%29+=+1 ----- Applying change of base to base 4
log+%284%2C+%28x+%2B+4%29%29+-+log+%284%2C+%28x%29%29%2F2+=+1
2%2Alog+%284%2C+%28x+%2B+4%29%29+-+log+%284%2C+%28x%29%29+=+2 ---- Multiplying by LCD, 2
log+%284%2C+%28x+%2B+4%29%29%5E2+-+log+%284%2C+%28x%29%29+=+2
log+%284%2C+%28x+%2B+4%29%5E2%2Fx%29+=+2
%28x+%2B+4%29%5E2%2Fx+=+4%5E2
%28x+%2B+4%29%5E2+=+16x ----- Cross-multiplying
x%5E2+%2B+8x+%2B+16+=+16x
x%5E2+%2B+8x+-+16x+%2B+16+=+0
x%5E2+-+8x+%2B+16+=+0


(x - 4)(x + 4) = 0

highlight_green%28x+=+4%29

======
Check
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log+%284%2C+%284+%2B+4%29%29+-+log+%2816%2C+%284%29%29+=+1
log+%284%2C+%288%29%29+-+log+%2816%2C+%284%29%29+=+1
1.5 - .5 = 1
1 = 1 (TRUE)

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