SOLUTION: Please help not sure how to solve this equation. (16^(x+2))/(64)=((1)/(64))^(3x-2)

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Question 1079943: Please help not sure how to solve this equation.
(16^(x+2))/(64)=((1)/(64))^(3x-2)
Found 3 solutions by Theo, MathLover1, MathTherapy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
your solution should be x = .4545454545 which is equivalent to 5/11.
if you graph y = 16^(x+2)/64 and y = (1/64)^(3x-2), their intersection should show you the solution.

the graph looks like this:

$$$

the calculations are messy so i'll write them manually and show you them in the following picture.

$$$

the rules of logarithms used are:

log(a^b) = b * log(a)

log(a * b) = log(a) + log(b)

note that:

log(a * b^c) = log(a) + log(b^c) = log(a) + c * log(b)

Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!


.........cross multiply



...replace as

















Answer by MathTherapy(10556)   (Show Source): You can put this solution on YOUR website!
Please help not sure how to solve this equation.
(16^(x+2))/(64)=((1)/(64))^(3x-2)


 ------- Converting all bases to base 4

 



2x + 1 = - 9x + 6 ------ Bases are equal and so are the exponents

2x + 9x = 6 - 1

11x = 5



It'as SIMPLE as that!!

There's ABSOLUTELY NO reason, at all, to use logs!! Doesn't make sense to me why someone would! 

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