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Solve for x: 27^5x−7=(1/81)^8x−3
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\[(3^{3})^{5x-1}=(3^{-4})^{8x-3}\] Make the base equal to each other so you can solve for the exponents themselves.. \[3^{15x-3}=3^{-32x+12}\] \[15x-3=32x+12\] Now just solve for x...
lindsayp024 used a 1 instead of the 7 in the original problem, but you should be able to see how to fix it.
oops sorry. I must have been drunk or something but the Idea of the problem is basically the same... make the same base and you should be able to make the problems equal to eachother
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