document.write( "Question 223055: log 16/log 32 \n" ); document.write( "

Algebra.Com's Answer #167032 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
There's a couple of ways to solve this:

Recognize that this is an example of the base conversion formula for logarithms, $\"log%28a%2C+%28x%29%29+=+log%28b%2C+%28x%29%29%2Flog%28b%2C+%28a%29%29\"$ with a = 32 and b = 10. So $\"log%28%2816%29%29%2Flog%28%2832%29%29+=+log%2832%2C+%2816%29%29\"$ . Let's call the later log \"x\": $\"x+=+log%2832%2C+%2816%29%29\"$ . Now let's rewrite this in exponential form: $\"32%5Ex+=+16\"$ . Next, since both 32 and 16 are both powers of 2, we'll rewrite them this way so we can have the same base: $\"%282%5E5%29%5Ex+=+2%5E4\"$ . Using a rule for exponents on the left: $\"2%5E%285x%29+=+2%5E4\"$ . This tells us that 5x = 4 or x = 4/5.
Let $\"x+=+log%28%2816%29%29%2Flog%28%2832%29%29\"$ . Multiply both sides by $\"log%28%2832%29%29\"$ : $\"x%2Alog%28%2832%29%29+=+log%28%2816%29%29\"$ . Using a rule for logarithms we can \"move\" the x from in front into the argument as an exponent: $\"log%28%2832%5Ex%29%29+=+log%28%2816%29%29\"$ . Since the log of $\"32%5Ex\"$ equals the log of 16, $\"32%5Ex+=+16\"$ . (We reached the same equation above, via a different route. The rest of the solution is the same.)

\n" ); document.write( "So, either way, $\"log%28%2816%29%29%2Flog%28%2832%29%29+=+4%2F5\"$ . \n" ); document.write( "

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