Question 253896
Unfortunately, you can't solve {{{27^(2x+1) = 94x}}} exactly since the variable is both in the exponent and outside the exponent. So you'll have to find the approximate solution.



If on the other hand the problem is {{{27^(2x+1) = 9^(4x)}}}, then...



{{{27^(2x+1) = 9^(4x)}}} Start with the given equation.



{{{(3^3)^(2x+1) = (3^2)^(4x)}}} Rewrite 27 as {{{3^3}}} and 9 as {{{3^2}}}



{{{3^(3(2x+1)) = 3^(2(4x))}}} Multiply the exponents.



{{{3^(3(2x+1)) = 3^(8x)}}} Multiply



{{{3(2x+1)=8x}}} Since the bases are equal, this means that the exponents are equal.



{{{6x+3=8x}}} Distribute.



{{{6x=8x-3}}} Subtract {{{3}}} from both sides.



{{{6x-8x=-3}}} Subtract {{{8x}}} from both sides.



{{{-2x=-3}}} Combine like terms on the left side.



{{{x=(-3)/(-2)}}} Divide both sides by {{{-2}}} to isolate {{{x}}}.



{{{x=3/2}}} Reduce.



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Answer:


So the solution is {{{x=3/2}}}