SOLUTION: Solve this equation. 5^x^3 = 25^x x =___________ ____(smallest value) x = x =___________ ______(largest value

Algebra ->  Test -> SOLUTION: Solve this equation. 5^x^3 = 25^x x =___________ ____(smallest value) x = x =___________ ______(largest value      Log On


   



Question 116625: Solve this equation.
5^x^3 = 25^x
x =________________(smallest value)
x =
x =__________________(largest value

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
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%5Ex is equal to x%2Alog+7
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In this problem you are given the equation:
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5%5E%28x%5E3%29+=+25%5Ex
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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%2810%2C+5%5E%28x%5E3%29%29+=+log%2810%2C25%5Ex%29
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bringing out the exponents and making them multipliers converts the problem to:
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x%5E3%2Alog%2810%2C5%29+=+x%2Alog%2810%2C25%29
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But notice on the right side that 25 can be replaced by 5%5E2 making the equation become:
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x%5E3%2Alog%2810%2C5%29+=+x%2Alog%2810%2C5%5E2%29
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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%5E3%2Alog%2810%2C5%29+=+2x%2Alog%2810%2C5%29
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Notice now that log%2810%2C5%29 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%2810%2C5%29 the logarithmic term just
cancels on both sides and the equation is reduced to:
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x%5E3+=+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%5E3+-+2x+=+0
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Factor the left side and you get:
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x%2A%28x%5E2+-2%29+=+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%5E2+-+2+=+0
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Add 2 to both sides and you get:
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x%5E2+=+2
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Solve for x by taking the square root of both sides and you have:
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x+=+%2Bsqrt%282%29 and x+=+-sqrt%282%29
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In summary, the three answers to this problem are:
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x+=+-sqrt%282%29
x+=+0
x+=+sqrt%282%29
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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%5E%28x%5E3%29+=+25%5Ex
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Substitute 0 for x and you have:
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5%5E0+=+25%5E0
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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%282%29 for x.
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The original equation is:
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5%5E%28x%5E3%29+=+25%5Ex
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Substitute sqrt%282%29%29 for x and get:
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5%5E%28%28sqrt%282%29%29%5E3%29+=+25%5E%28sqrt%282%29%29%29
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On the right side replace 25 with 5%5E2 and the equation becomes:
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5%5E%28%28sqrt%282%29%29%5E3%29+=+%285%5E2%29%5E%28sqrt%282%29%29
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On the left side cubing sqrt%282%29 is equivalent to %28sqrt%282%29%29%5E2%2Asqrt%282%29 and %28sqrt%282%29%29%5E2+=+2
so cubing sqrt%282%29 results in 2%2Asqrt%282%29. This makes the equation become:
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5%5E%282%2Asqrt%282%29%29+=+%285%5E2%29%5E%28sqrt%282%29%29
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Finally on the right side, the power rule of exponents tells you to multiply 2 times sqrt%282%29
and this makes the equation become:
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5%5E%282%2Asqrt%282%29%29+=+5%5E%282%2Asqrt%282%29%29
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Note that the left side now equals the right side, so x+=+sqrt%282%29 is a good solution
to the problem.
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You can do the proof for x+=+-sqrt%282%29. It works very similar to the one for x+=+sqrt%282%29.
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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 ...