document.write( "Question 421639: log base8 2x^2 + log base8 4 = 5 \n" ); document.write( "
Algebra.Com's Answer #294483 by richard1234(7193)\"\" \"About 
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We have \"log%288%2C+2x%5E2%29+%2B+log%288%2C+4%29+=+5\". We can use the formula \"log%28a%2Cx%29+%2B+log%28a%2Cy%29+=+log%28a%2Cxy%29\" to obtain\r
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\n" ); document.write( "\n" ); document.write( "\"log%288%2C+8x%5E2%29+=+5\"\r
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\n" ); document.write( "\n" ); document.write( "Raise each side to the exponent with base 8, i.e. \"8%5E%28log%288%2C+8x%5E2%29%29+=+8%5E5\" to get\r
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\n" ); document.write( "\n" ); document.write( "\"8x%5E2+=+8%5E5\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2+=+8%5E4\"\r
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\n" ); document.write( "\n" ); document.write( "x = 64 or -64. We have to be especially careful dealing with negative solutions or obtaining a negative in the logarithm, since the logarithm is not defined on negative numbers (disregarding complex values). We can check that both solutions satisfy.
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