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

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Please help not sure how to solve this equation. (16^(x+2))/(64)=((1)/(64))^(3x-2)      Log On


   

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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
%2816%5E%28x%2B2%29%29%2F64=%281%2F64%29%5E%283x-2%29+

%2816%5E%28x%2B2%29%29%2F64=%281%5E%283x-2%29%29%2F%2864%5E%283x-2%29+%29.........cross multiply

%2816%5E%28x%2B2%29%29%2864%5E%283x-2%29+%29=64%281%5E%283x-2%29%29

%28%282%5E4%29%5E%28x%2B2%29%29%28%282%5E6%29%5E%283x-2%29+%29=2%5E6%281%5E%283x-2%29%29...replace 1 as 2%5E0



2%5E%284x%2B8%29%2A2%5E%286%283x-2%29%29=2%5E6%2A2%5E%280%283x-2%29%29

2%5E%284x%2B8%29%2A2%5E%2818x-12%29+=2%5E6%2A2%5E0

2%5E%284x%2B8%2B18x-12%29+=2%5E6%2A1

2%5E%2822x-4%29+=2%5E6

22x-4+=6
22x+=6%2B4
22x+=10
x+=10%2F22
x+=5%2F11

Answer by MathTherapy(10556) About Me  (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)
16%5E%28x+%2B+2%29%2F64+=+1%2F64%5E%283x+-+2%29

%284%5E2%29%5E%28x+%2B+2%29%2F4%5E3+=+1%2F%284%5E3%29%5E%283x+-+2%29 ------- Converting all bases to base 4

4%5E%282x+%2B+4+-+3%29+=+1%2F4%5E%289x+-+6%29 

4%5E%282x+%2B+1%29+=+4%5E%28-+%289x+-+6%29%29

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

2x + 9x = 6 - 1

11x = 5

highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+5%2F11%29%29

It'as SIMPLE as that!!

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