SOLUTION: Solve 16^(x+3)=64^(2x-5)

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Question 896967: Solve 16^(x+3)=64^(2x-5)
Found 2 solutions by thesvw, MathTherapy:
Answer by thesvw(77) About Me  (Show Source):
You can put this solution on YOUR website!
16%5E%28x%2B3%29=64%5E%282x-5%29
Take log to the base 4.
log%2816%5E%28x%2B3%29%29=log%2864%5E%282x-5%29%29
%28x%2B3%29%2Alog16=+%282x-5%29log64
log is to the base 4. So you can write..
%28x%2B3%29%2A2=+%282x-5%293
2x+%2B6+=+6x+-15
21=4x
x=21%2F4

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Solve 16^(x+3)=64^(2x-5)


16%5E%28x+%2B+3%29+=+64%5E%282x+-+5%29

%284%5E2%29%5E%28x+%2B+3%29+=+%284%5E3%29%5E%282x+-+5%29

4%5E%282%28x+%2B+3%29%29+=+4%5E%283%282x+-+5%29%29

2(x + 3) = 3(2x - 5) ------ Bases are equal, and so are their exponents

2x + 6 = 6x - 15

2x - 6x = - 15 - 6

- 4x = - 21

x+=+%28-+21%29%2F%28-+4%29, or highlight_green%28highlight_green%285.25%29%29

You can do the check!! 



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