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

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve. log(base4)(x+4)-log(base16)x=1      Log On


   

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(10555) About Me  (Show Source):
You can put this solution on YOUR website!

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

log_%5B4%5D+%28x+%2B+4%29+-+log_%5B16%5D+%28x%29+=+1
log_%5B4%5D+%28x+%2B+4%29+-+log_%5B4%5D+x%2Flog_%5B4%5D+16+=+1 ----- Applying change of base to base 4
log_%5B4%5D+%28x+%2B+4%29+-+%28log_%5B4%5Dx%29%2F2+=+1
2%2Alog_%5B4%5D+%28x+%2B+4%29+-+log_%5B4%5D+x+=+2 ----- Multiplying by LCD, 2
log_%5B4%5D+%28x+%2B+4%29%5E2+-+log_%5B4%5D+x+=+2
log_%5B4%5D+%28%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
======
log_%5B4%5D+%284+%2B+4%29+-+log_%5B16%5D+%284%29+=+1
log_%5B4%5D+8+-+log_%5B16%5D+4+=+1
1.5 - .5 = 1
1 = 1 (TRUE)

Send comments and “thank-yous” to “D” at MathMadEzy@aol.com