SOLUTION: sqrt(125^x) = 5/(25^x)

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Question 1148840: sqrt(125^x) = 5/(25^x)
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
start with:
sqrt(125^x) = 5/(25^x)
rewrite it as:
(125^x)^(1/2) = 5/(25^x)
simplify that to:
125^(x/2) = 5/(25^x)
multiply both sides of that equation by 25^x to get:
125^(x/2) * 25^x = 5
take the log of both sides of this equation to get:
log(125^(x/2) * 25^x) = log(5)
use log rules to translate that to:
log(125^(x/2)) + log(25^x) = log(5)
simplify that to:
x/2 * log(125) + x * log(25) = log(5)
factor out the x to get:
x * (1/2 * log(125) + log(25)) = log(5)
divide both sides of that equation by (1/2 * log(125) + log(25)) to get:
x = log(5) / (1/2 * log(125) + log(25))
solve for x to get:
x = .2857142857
replace x in the original equation to get:
sqrt(125^x) = 5/(25^x) becomes sqrt(125^.2857142857) = 5 / 25^.2857142857 which becomes 1.993235316 = 1.993235316 which is true.

your solution is that the value for x is .2857142857

the properties of logs that were used are:

log(a * b) = log(a) + log(b) and log(a^x) = x*log(a)


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt(125^x) = 5/(25^x)
matrix%281%2C3%2C+sqrt%28125%5Ex%29%2C+%22=%22%2C+5%2F%2825%5Ex%29%29
matrix%281%2C3%2C+%28sqrt%28125%5Ex%29%29%5E2%2C+%22=%22%2C+%285%2F25%5Ex%29%5E2%29 --- Squaring both sides
matrix%281%2C3%2C+125%5Ex%2C+%22=%22%2C+5%5E2%2F25%5E%282x%29%29 
matrix%281%2C3%2C+%285%5E3%29%5Ex%2C+%22=%22%2C+5%5E2%2F%285%5E2%29%5E%282x%29%29 ------- Changing bases
matrix%281%2C3%2C+5%5E%283x%29%2C+%22=%22%2C+5%5E2%2F5%5E%284x%29%29 
matrix%281%2C3%2C+5%5E%283x%29%2C+%22=%22%2C+5%5E%282+-+4x%29%29 
3x = 2 - 4x ------ Equating exponents, since their bases are equal
3x + 4x = 2
7x = 2
highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+2%2F7%29%29

FYI: LOGS are definitely NOT REQUIRED to solve this, as it's not that much of a COMPLEX exponents-problem.