document.write( "Question 240762: Solve for b:\r
\n" ); document.write( "\n" ); document.write( "logb8=-3 \n" ); document.write( "
Algebra.Com's Answer #176374 by jsmallt9(3759)\"\" \"About 
You can put this solution on YOUR website!
When the variable is in the argument or the base of a logarithm, you solve for that variable by rewriting the equation in exponential form. This is done by using the fact that \"log%28a%2C+%28p%29%29+=+q\" is equivalent to \"a%5Eq+=+p\".

\n" ); document.write( "Rewriting your equation in exponential form we get:
\n" ); document.write( "\"b%5E%28-3%29+=+8\"
\n" ); document.write( "Now we have an equation we can solve. There are several ways we could solve this. One would be to raise both sides to the -1/3 power. (You'll see wht when we're done)
\n" ); document.write( "\"%28b%5E%28-3%29%29%5E%28-1%2F3%29+=+8%5E%28-1%2F3%29\"
\n" ); document.write( "which results in:
\n" ); document.write( "\"b+=++8%5E%28-1%2F3%29\"
\n" ); document.write( "We have been trying to solve for b and, as you can see, we have been able to do this in one step (raising to the -1/3 power). All that is left is to simpilfy the right side. To do this by hand it helps to factor the exponent:
\n" ); document.write( "\"b+=++8%5E%28%281%2F3%29%2A%28-1%29%29\"
\n" ); document.write( "or
\n" ); document.write( "\"b+=+%288%5E%281%2F3%29%29%5E%28-1%29\"
\n" ); document.write( "Since 1/3 as an exponent means \"cube root of\" and the cube root of 8 is 2:
\n" ); document.write( "\"b+=+2%5E%28-1%29\"
\n" ); document.write( "Since -1 as an exponent means \"reciprocal of\" and the reciprocal of 2 is 1/2:
\n" ); document.write( "\"b+=+1%2F2\" \n" ); document.write( "
\n" );