SOLUTION: 4^log3 base 9 + 9^log 4 base 2 = 10^log 83 base x, find x

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Question 970602: 4^log3 base 9 + 9^log 4 base 2 = 10^log 83 base x, find x
Found 2 solutions by khai, MathTherapy:
Answer by khai(18) About Me  (Show Source):
You can put this solution on YOUR website!
let 4^log3 base 9 =y {solve for y}
4^y = log3 base 9
2^(2y) = 2^-1 --------- {0.5 = 2^-}
2y = -1 ---- {same base}
y = -1/2
let 9^log4 base 2 = t {solve for t}
9^t = log4 base 2
9^t = 2
log2 base 9 = t
t = 0.3154648768
add the first two, you will get
-0.1845351232 = 10^log 83 base x
10^(-0.1845351232) = log 83 base x
0.6538300499 = log 83 base x
0.6538300499 = (log 83 base 10)/(log x base 10)
(log 83 base 10)/(0.6538300499) = (log x base 10)
2.935132903 = (log x base 10)
10^(2.935132903)= x
x = 861.2572741

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
4^log3 base 9 + 9^log 4 base 2 = 10^log 83 base x, find x
If 4%5E%28log+%289%2C+3%29%29+%2B+9%5E%28log+%282%2C+4%29%29+=+10%5E%28log+%28x%2C+83%29%29, then: highlight_green%28x+=+10%29