Question 70853
{{{(3^(5x))*(9^(x^2))=27}}}
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Interesting problem.
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First recognize that {{{9^(x^2)}}} can be written as {{{(3*3)^(x^2)}}}. By the rules of
exponents this can be expanded to {{{3^(x^2)*3^(x^2)}}} and this multiplication can be
done by adding the exponents to get {{{3^(2x^2)}}}.  We can now replace {{{9^(x^2)}}}
by {{{3^(2x^2)}}}.  So let's do that very thing.  When the substitution is done, the
problem becomes:
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{{{(3^(5x))* 3^(2x^2)=27}}}
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Since these two terms have the common base of 3, the multiplication can be done by adding
the exponents to get:
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{{{3^(2x^2+5x) = 27}}}
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Now let's take the log of both sides, and the equation becomes:
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{{{log(3^(2x^2+5x))=log(27)}}}
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On the left side you can take the exponent away and make it the multiplier of the log term:
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{{{(2x^2+5x)*log(3) = log(27)}}}
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But on the right side the 27 can be replaced by {{{3^3}}} and the equation becomes:
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{{{(2x^2+5x)*log(3) = log(3^3)}}}
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And now on the right side we can take the exponent out and use it as the multiplier of 
the log to make the equation:
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{{{(2x^2+5x)*log(3) = 3*log(3)}}}
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Note that the log(3) is a factor on both sides.  If we divide both sides by log(3)
it cancels out and we are left with the "nice" equation:
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{{{2x^2 + 5x = 3}}}
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Subtract 3 from both sides:
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{{{2x^2 + 5x - 3 = 0}}}
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The left side factors to give:
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{{{(2x -1)*(x+3) = 0}}}
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Note that if either of the two factors equals 0, the equation will be true. So we can
find the values of x by setting the factors equal to zero.
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{{{2x-1 = 0}}}
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Add 1 to both sides and then divide by 2 to find that {{{x=1/2}}} is a solution.
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Now set the other factor equal to zero.
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{{{x+3=0}}} 
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Subtract 3 from both sides and you have: {{{x= -3}}}
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So 1/2 and -3 are the two values of x.  If you plug them, one at a time, into the equation 
you were given as the problem and work out the resulting equation using a calculator, 
you will find that they both work.
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Like I said at the start, "Interesting problem."  Good practice too. I hope you can work
your way through all the steps.