document.write( "Question 171558: log12^x=1
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Algebra.Com's Answer #126711 by jim_thompson5910(35256)\"\" \"About 
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\"log%2812%2C%28x%29%29=1\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"12%5E1=x\" Rewrite the equation using the property: \"log%28b%2C%28x%29%29=y\" ====> \"b%5Ey=x\"\r
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\n" ); document.write( "\n" ); document.write( "\"12=x\" Raise 12 to the first power to get 12\r
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\n" ); document.write( "\n" ); document.write( "So the answer is \"x=12\"
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