Question 851467: SOLVE FOR THE VAULE OF X: 27^X=9^X+2
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the equation, as far as i can tell is:
27^x = 9^(x+2)
if so, then the answer is x = 4 which has been derived as shown below:
start with:
27^x = 9^(x+2)
take the log of both sides of the equation to get:
log(27^x) = log(9^(x+2))
use the property of log(a^x) = x*log(a) to get:
x*log(27) = (x+2)*log(9)
simplify by removing parentheses to get:
x*log(27) = x*log(9) + 2*log(9)
subtract x*log(9) from both sides of the equation to get:
x*log(27) - x*log(9) = 2*log(9)
factor out the x to get:
x*(log(27) - log(9)) = 2*log(9)
divide both sides of the equation by (log(27)-log(9)) to get:
x = 2*log(9) / (log(27) - log(9))
solve for x by taking the log of 9 and log of 27 and evaluating the equation to get:
x = 4
replace x in the original equation to get:
27^x = 9^(x+2) becomes 27^4 = 9^6 which becomes 531441 = 531441
this confirms the solution is good.
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