document.write( "Question 710345: Solve the expression \"Ta%5Et+=+Qk%5Et\" for t assuming that all other letters are positive constants. \n" ); document.write( "
Algebra.Com's Answer #437283 by jsmallt9(3759)\"\" \"About 
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\"Ta%5Et+=+Qk%5Et\"
\n" ); document.write( "[I'm assuming that the upper case tee, T, is not the same variable as the lower case tee, t, we are solving for.]

\n" ); document.write( "Solving for a variable which is in an exponent usually involves logarithms. (An exception would be when the equation can be rewritten so that each side of the equation is a power of the same number. Your equation cannot be rewritten this way.)

\n" ); document.write( "We can start by finding the logarithm of each side. The base we choose for the logarithm does not matter in any important way. But if we choose a base for the logarithm that matches the base of an exponent we will get a simpler expression for an answer. So we will choose base \"a\" logarithms which matches the base of the first exponent of t:
\n" ); document.write( "\"log%28a%2C+%28Ta%5Et%29%29+=+log%28a%2C+%28Qk%5Et%29%29\"
\n" ); document.write( "The reason we use logarithms in the first place is that they have a property, \"log%28m%2C+%28p%5En%29%29+=+n%2Alog%28m%2C+%28p%29%29\", which allows us to move the exponent of the argument (where the variable we are solving for is) out in front of the log (where are can \"get at it\" with regular Algebra). Before we can use this property, however, the argument has to be just a base and an exponent.

\n" ); document.write( "So before we move the exponent we must find a way to separate out the \"T\" and the \"Q\" from the base an its exponent. Fortunately there is another property of logarithms, \"log%28m%2C+%28p%2Aq%29%29+=+log%28m%2C+%28p%29%29+%2B+log%28m%2C+%28q%29%29\", which allows us to separate the factors of the argument of a log into separate logs. Using this property on both sides of the equation we get:
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\n" ); document.write( "Now we can use the other property to move the exponents of the arguments of the second log on each side of the equation:
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\n" ); document.write( "Whenever the base of a log matches the argument of that log, the log is equal to 1. So the second log on the left side is 1. (This is why matching the base of the log to the base of the exponent results in a simpler expression.)
\n" ); document.write( "\"log%28a%2C+%28T%29%29+%2B+t%2A1+=+log%28a%2C+%28Q%29%29+%2B+t%2Alog%28a%2C+%28k%29%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"log%28a%2C+%28T%29%29+%2B+t+=+log%28a%2C+%28Q%29%29+%2B+t%2Alog%28a%2C+%28k%29%29\"

\n" ); document.write( "Now that the t's are out of the exponents we can solve for it. First let's gather the terms with t on one side of the equation and the other terms on the other side. Subtracting t and \"log%28a%2C+%28Q%29%29\" form each side we get:
\n" ); document.write( "\"log%28a%2C+%28T%29%29+-+log%28a%2C+%28Q%29%29+=+t%2Alog%28a%2C+%28k%29%29+-+t\"
\n" ); document.write( "The t terms are not like terms so we cannot subtract them. But we can factor out t:
\n" ); document.write( "\"log%28a%2C+%28T%29%29+-+log%28a%2C+%28Q%29%29+=+t%2A%28log%28a%2C+%28k%29%29+-+1%29\"
\n" ); document.write( "And then divide both sides by \"%28log%28a%2C+%28k%29%29+-+1%29\":
\n" ); document.write( "\"%28log%28a%2C+%28T%29%29+-+log%28a%2C+%28Q%29%29%29%2F%28log%28a%2C+%28k%29%29+-+1%29+=+t\"
\n" ); document.write( "This is the equation solved for t.

\n" ); document.write( "Notes:
  • When we found the log of each side we could have used any base for the logarithm. If we had chosen a different base then our answer would look different from what we found above. These different-looking answers (assuming no mistakes were made) would both be correct. And since there are an infinite number of possible bases to logarithms there will be an infinite number of different-looking but correct answers to this problem!
  • The numerator of our answer fits the pattern of one side of yet another property of logarithms:
    \n" ); document.write( "\"log%28m%2C+%28p%2Fq%29%29+=+log%28m%2C+%28p%29%29+-+log%28m%2C+%28q%29%29\"
    \n" ); document.write( "So that numerator could be rewritten:
    \n" ); document.write( "\"%28log%28a%2C+%28T%2FQ%29%29%29%2F%28log%28a%2C+%28k%29%29+-+1%29+=+t\"
    \n" ); document.write( "This may or may not be a preferred form for the answer we found above.
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