Question 116625
In case you can still use an answer to this problem, here's how you can do it.
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The biggest lesson that you can learn from this problem is that anytime you see a variable used
as an exponent, you should think of the possibility of using logarithms to solve the problem
because if you take the logarithm of such a quantity, the exponent can be brought out as the
multiplier of the logarithm. This is true regardless of base in the logarithm. For example:
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{{{log 7^x}}} is equal to {{{x*log 7}}}
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In this problem you are given the equation:
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{{{5^(x^3) = 25^x}}}
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If you take the Logarithm of both sides (any base will do, but just for simplicity let's
use base 10), the equation becomes:
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{{{log(10, 5^(x^3)) = log(10,25^x)}}}
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bringing out the exponents and making them multipliers converts the problem to:
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{{{x^3*log(10,5) = x*log(10,25)}}}
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But notice on the right side that 25 can be replaced by {{{5^2}}} making the equation become:
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{{{x^3*log(10,5) = x*log(10,5^2)}}}
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Applying the exponent property again on the right side, the 2 comes out as a multiplier and
you have:
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{{{x^3*log(10,5) = 2x*log(10,5)}}}
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Notice now that {{{log(10,5)}}} is a factor of both sides, and it is just a number that you
can get using a calculator. [Should you be interested, that number is 0.698970004].
If both sides of the equation are divided by {{{log(10,5)}}} the logarithmic term just
cancels on both sides and the equation is reduced to:
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{{{x^3 = 2x}}}
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Get rid of the 2x on the right side by subtracting 2x from both sides to get:
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{{{x^3 - 2x = 0}}}
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Factor the left side and you get:
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{{{x*(x^2 -2) = 0}}}
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This equation will be true if either of the factors on the left side equals zero because then
the left side would involve a multiplication by zero and that zero multiplier would make
the entire left side equal to zero ... therefore equaling the right side.
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So setting the first factor equal to zero just gives:
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{{{x = 0}}}
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Then setting the second factor equal to zero gives:
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{{{x^2 - 2 = 0}}}
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Add 2 to both sides and you get:
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{{{x^2 = 2}}}
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Solve for x by taking the square root of both sides and you have:
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{{{x = +sqrt(2)}}} and {{{x = -sqrt(2)}}}
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In summary, the three answers to this problem are: 
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{{{x = -sqrt(2)}}}
{{{x = 0}}}
{{{x = sqrt(2)}}}
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If you substitute each of these three values back into the original equation, you will find
that the equation is true.
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For example: Start with the original equation:
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{{{5^(x^3) = 25^x}}}
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Substitute 0 for x and you have:
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{{{5^0 = 25^0}}}
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but you know that that when any quantity is raised to the zero power the answer is 1. So this
equation reduces to:
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{{{ 1 = 1}}}
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And this is obviously true ... so x = 0 is a good solution.
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Next start with the original equation and substitute {{{sqrt(2)}}} for x. 
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The original equation is:
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{{{5^(x^3) = 25^x}}}
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Substitute {{{sqrt(2))}}} for x and get:
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{{{5^((sqrt(2))^3) = 25^(sqrt(2)))}}}
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On the right side replace 25 with {{{5^2}}} and the equation becomes:
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{{{5^((sqrt(2))^3) = (5^2)^(sqrt(2))}}}
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On the left side cubing {{{sqrt(2)}}} is equivalent to {{{(sqrt(2))^2*sqrt(2)}}} and {{{(sqrt(2))^2 = 2}}}
so cubing {{{sqrt(2)}}} results in {{{2*sqrt(2)}}}. This makes the equation become:
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{{{5^(2*sqrt(2)) = (5^2)^(sqrt(2))}}}
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Finally on the right side, the power rule of exponents tells you to multiply 2 times {{{sqrt(2)}}}
and this makes the equation become:
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{{{5^(2*sqrt(2)) = 5^(2*sqrt(2))}}}
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Note that the left side now equals the right side, so {{{x = sqrt(2)}}} is a good solution
to the problem. 
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You can do the proof for {{{x = -sqrt(2)}}}. It works very similar to the one for {{{x = sqrt(2)}}}.
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I know this is late, but the lessons involved in solving this problem are hopefully in
time to help you understand the concepts involved in getting a solution.
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Happy holidays to you and yours ...