Question 930561
3^5x+2/9^1-x=27^4+3x/729
<pre><font face = "Tohoma" size = 4 color = "indigo"><b>Could be: {{{highlight_green(x = - 3)}}}. 
Then again, it might not be. Do you know why or why not?

Here's the solution you requested:
{{{(3^(5x + 2))/9^(1 - x) = 27^(4 + 3x)/729}}}
{{{(3^(5x + 2))/(3^2)^(1 - x) = (3^3)^(4 + 3x)/3^6}}}
{{{(3^(5x + 2))/3^(2(1 - x)) = 3^(3(4 + 3x))/3^6}}}
{{{(3^(5x + 2))/3^(2 - 2x) = 3^(12 + 9x)/3^6}}}
{{{3^((5x + 2) - (2 - 2x)) = 3^(12 + 9x - 6)}}}
{{{5x + 2 - 2 + 2x = 12 + 9x - 6}}} ------- Bases are equivalent and so are their exponents
{{{5x + 2x + 2 - 2 = 6 + 9x}}}
{{{7x = 6 + 9x}}}
{{{7x - 9x = 6}}}
{{{- 2x = 6}}}
{{{x = 6/(- 2)}}}, or {{{highlight_green(x = - 3)}}}
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</font face = "Tohoma" size = 4 color = "indigo"></b>Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.
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