document.write( "Question 706888: 16^(log4(3x))\r
\n" ); document.write( "\n" ); document.write( "I need to simplify the expression and show all the steps.\r
\n" ); document.write( "\n" ); document.write( "Thank you so much! \n" ); document.write( "
Algebra.Com's Answer #435514 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"16%5E%28log%284%2C%283x%29%29%29\"
\n" ); document.write( "The exponent's base is 16 and the logarithm's base is 4. It will make things much easier if we can change one of these bases into the other or change both bases into some third number.

\n" ); document.write( "With bases of 16 and 4 it should be easy to change one of them into the other. I'm going to change the 16 into a 4. (I'll also show you changing the 4 into a 16 later.) Since \"16+=+4%5E2\":
\n" ); document.write( "\"%284%5E2%29%5E%28log%284%2C%283x%29%29%29\"
\n" ); document.write( "(Note my use a parentheses. It is an extremely good habit to use parentheses like this when making substitutions.) Our expression is now a power of a power. The rule for these exponents is to multiply them:
\n" ); document.write( "\"4%5E%282%2Alog%284%2C%283x%29%29%29\"
\n" ); document.write( "If we can get the 2 out of the way (so the expression is 4 to a base4 log power) we will be almost done. Fortunately there is a property of logarithms, \"k%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Ek%29%29\", which allows us to move the coefficient of a log into the argument as its exponent. Using the property we can move the 2:
\n" ); document.write( "\"4%5E%28log%284%2C%28%283x%29%5E2%29%29%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"4%5E%28log%284%2C%289x%5E2%29%29%29\"
\n" ); document.write( "We should know that that this simplifies to:
\n" ); document.write( "\"9x%5E2\"
\n" ); document.write( "(After all, \"log%284%2C+%289x%5E2%29%29\" represents the exponent one would put on a 4 to get a result of \"9x%5E2\". And look at where it is! It's the exponent on a 4! So of course we get \"9x%5E2\"!).

\n" ); document.write( "P.S. Changing the base of the logarithm into 16 (instead of the 16 into a 4):
\n" ); document.write( "\"16%5E%28log%284%2C%283x%29%29%29\"
\n" ); document.write( "With the change of base formula, \"log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29\", we can change the base of the logarithm:
\n" ); document.write( "\"16%5E%28%28%28log%2816%2C%283x%29%29%29%2Flog%2816%2C+%284%29%29%29%29\" (In case you can't see it, the numerator of the exponent is: \"log%2816%2C+%283x%29%29\")
\n" ); document.write( "The log in the denominator represents the exponent one would put on a 16 to get a 4. Since 4 is the square root of 16 and since square root is 1/2 as an exponent, the log in the denominator is 1/2:
\n" ); document.write( "\"16%5E%28%28%28log%2816%2C%283x%29%29%29%2F%28%281%2F2%29%29%29%29\" (In case you can't see it, the numerator of the exponent is: \"log%2816%2C+%283x%29%29\")
\n" ); document.write( "And since dividing by 1/2 is the same as multiplying by 2 this simplifies to:
\n" ); document.write( "\"16%5E%282%2Alog%2816%2C%283x%29%29%29\"
\n" ); document.write( "Again we use the property of logs to move the 2 out of the way:
\n" ); document.write( "\"16%5E%28log%2816%2C%28%283x%29%5E2%29%29%29\"
\n" ); document.write( "which simplifies to
\n" ); document.write( "\"16%5E%28log%2816%2C%289x%5E2%29%29%29\"
\n" ); document.write( "which simplifies (for the same reason as our earlier solution) to:
\n" ); document.write( "\"9x%5E2\"

\n" ); document.write( "P.P.S. Since 16 and 4 are both powers of 2 you could also solve this problem by changing both 16 and 4 into 2's:
\n" ); document.write( "\"16%5E%28log%284%2C%283x%29%29%29\"
\n" ); document.write( "\"%282%5E4%29%5E%28%28log%282%2C%283x%29%29%29%2Flog%282%2C+%284%29%29%29\" (In case you can't see it, the numerator of the outer exponent is: \"log%282%2C+%283x%29%29\")
\n" ); document.write( "But changing both bases to a third number like this should be your last resort. If it is possible to change one base into the other, it will always be easier that way than to change both bases into a third number. If you're curious I'll leave it up to finish this. I've already shown you the hardest step. \n" ); document.write( "
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