SOLUTION: 6*4^x=99

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Question 472238: 6*4^x=99
Found 2 solutions by Alan3354, bucky:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
6%2A4%5Ex=99
4%5Ex+=+16.5
x*log(4) = log(16.5)
etc

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
The major problem in this question is the fact that the unknown variable, x, is an exponent. Whenever you have an unknown variable as an exponent, you should think of using logarithms to solve for this variable.
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That being the case, we can begin by dividing both sides of this equation by 6 so that we have the term containing the exponent x all by itself on the left side. When we divide both sides by 6 the equation becomes:
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4%5Ex+=+99%2F6
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Next we can take the logarithm of both sides to get:
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+log4%5Ex+=+log%2899%2F6%29
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On the right side we should recognize that the logarithm of a number is just a number that can be looked up on a calculator. So we can begin by dividing 99 by 6 to get:
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99%2F6+=+16.5
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And substituting this into our problem changes the right side as follows:
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log4%5Ex+=+log+16.5
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Using a calculator, enter 16.5 and press the log key to find that the log of 16.5 is 1.217483944. Substitute this value as a replacement for the right side and the equation is then:
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log4%5Ex+=+1.217483944
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Now let's work on the left side. By the rules of logarithms, when we have an exponent in the number that the logarithm is operating on, the exponent can be brought out as a multiplier of the logarithm. So we can bring out the x to become a multiplier of the logarithm as follows:
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x%2Alog4+=+1.217483944
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Recognize that log+4+ is just a number that can be determined by using a calculator. Enter 4 and press the log key to find that log4+=+0.602059991. Substitute this value for log+4 in the equation and it becomes:
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x%2A0.602059991=1.217483944
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Solve for x by dividing both sides by 0.602059991 and we get:
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x+=1.217483944%2F0.602059991
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Do the division on the right side using a calculator and you find that:
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x+=+2.02219706
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That's the answer that we were looking for.
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Hope this helps you to understand what you can do to solve equations for which the variable is in an exponent and some of the things you can do with logarithms.
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