document.write( "Question 633354: Greetings I am having trouble with a log problem to solve for X\r
\n" ); document.write( "\n" ); document.write( "in logx32=5/3 \r
\n" ); document.write( "\n" ); document.write( "I thought I must change into exponential form and get x^5/3 =2^5
\n" ); document.write( "and then go into log form to get
\n" ); document.write( "3/5logx=5log2 and then I am lost.
\n" ); document.write( "please help? \n" ); document.write( "
Algebra.Com's Answer #398840 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"log%28x%2C+%2832%29%29=5%2F3+\"
\n" ); document.write( "You made a great first step with
\n" ); document.write( "\"x%5E%28%285%2F3%29%29+=+2%5E5\"
\n" ); document.write( "Not only did you get the variable out of the logarithm, but you recognized that \"32+=+2%5E5\" which turns out to be a very helpful.

\n" ); document.write( "But it's not a good idea to go back to log form. That just puts the x back in the logarithm which is not where we want it. Instead, what we want is to change the exponent on x to a 1. If we can do that, we'll be finished since \"x%5E1+=+x\"!

\n" ); document.write( "So how do we change an exponent? Well there are several rules for exponents that tell us: \"When we do __________, we change the exponents in this way.\". If we use the multiplication rule (add the exponents) to get an exponent of 1 (by multiplying by \"x%5E%28%28-2%2F3%29%29\" we could get the exponent of 1 on the left but then we would have \"x%5E%28%28-2%2F3%29%29\" on the the right side. If we use the division rule (subtract the exponents) we have a similar problem, We can get the exponent we want on the left but we end up with an x on the right side. The rule that will help us is the power of a power rule (multiply the exponents). If we raise both sides to the right power, we can get an exponent of 1 on the left and there will not be any x's on the right.

\n" ); document.write( "So what power to we raise \"x%5E%28%285%2F3%29%29\" to to get an exponent of 1? Knowing that we will be multiplying this exponent by 5/3 and knowing that multiplying reciprocals always results in 1, the right power to use is the reciprocal of 5/3: 3/5.
\n" ); document.write( "\"%28x%5E%28%285%2F3%29%29%29%5E%28%283%2F5%29%29+=+%282%5E5%29%5E%28%283%2F5%29%29\"
\n" ); document.write( "On the left, we get the 1 we wanted. On the right, after we multiply 5 and 3/5, we get:
\n" ); document.write( "\"x+=+2%5E3\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"x+=+8\"

\n" ); document.write( "And, as usual with problems where the variable was in the argument or base of a logarithm, we must check the solution to ensure that all arguments and bases of all logarithms remain valid (positive for both and bases cannot be a 1 either). If any argument or base ends up being invalid we must reject that \"solution\", even if it's the only one we found!

\n" ); document.write( "Use the original equation to check:
\n" ); document.write( "\"log%28x%2C+%2832%29%29=5%2F3+\"
\n" ); document.write( "Checking x = 8:
\n" ); document.write( "\"log%28%288%29%2C+%2832%29%29=5%2F3+\"
\n" ); document.write( "We can quickly see that the base is positive (and not a 1) and the argument is positive. So our solution checks! \n" ); document.write( "
\n" );