SOLUTION: Solve for x: {(3^x9^x+20)/(27^2x+5)}=81^x+5

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Question 74477: Solve for x: {(3^x9^x+20)/(27^2x+5)}=81^x+5
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
I have added parentheses to the equation which I think were missing:
(3^x*9^(x+20))/(27^(2x+5))=81^(x+5)

Reduce all the numbers in the unknown terms to powers of 3 and simplify
or %283%5E%283x%2B40%29%29%2F%283%5E%286x%2B15%29%29=3%5E%284x%2B20%29

Multiply both sides by the denominator on the left side
3%5E%283x%2B40%29=3%5E%2810x%2B35%29

Since bases are the same, exponents are equal
3x%2B40=10x%2B35 so x=5%2F7