document.write( "Question 1057131: I just need someone to explain this and why 2 is the answer:
\n" ); document.write( "3^(log base 9 (4))= 2\r
\n" ); document.write( "\n" ); document.write( "I'm not sure how to write out log base correctly, but it's log(4) with the base of 9.
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Algebra.Com's Answer #672208 by solve_for_x(190)\"\" \"About 
You can put this solution on YOUR website!
When you have something like log (base 9) 4, you can rewrite it as \"log+4+%2F+log+9\"\r
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\n" ); document.write( "\n" ); document.write( "And when you do that, you can use any base for the logarithms that you want.\r
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\n" ); document.write( "\n" ); document.write( "So, instead of using the common base-10 log, we can write is using a base-3 log:\r
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\n" ); document.write( "\n" ); document.write( "\"log9+%284%29+=+log+4+%2F+log+9+=+log3+%284%29+%2F+log3+%289%29\"\r
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\n" ); document.write( "\n" ); document.write( "But log3 (9) = 2:\r
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\n" ); document.write( "\n" ); document.write( "\"log9+%284%29+=+log3+%284%29+%2F+2\"\r
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\n" ); document.write( "\n" ); document.write( "Substituting \"log3%284%29%2F2\" in place of \"log9%284%29\" gives:\r
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\n" ); document.write( "\n" ); document.write( "\"3%5E%28log3%284%29%2F2%29\"\r
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\n" ); document.write( "\n" ); document.write( "which equals:\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%283%5E%28log3%284%29%29%29++\"\r
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\n" ); document.write( "\n" ); document.write( "And since \"3%5E%28log3%284%29%29+=+4+\", the expression is equal to:\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%284%29+=+2\"
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