document.write( "Question 192060: This is the first time my instructor has presented problems written this way. I'm clueless. I would appreciate if someone could show me how to do this one so I can handle the rest. Thank you
\n" ); document.write( "If a is a positive real number, such that a is not equal to 1, and ax = b,
\n" ); document.write( "then the logarithmic function is represented as:
\n" ); document.write( "A) log a^x b = x
\n" ); document.write( "B) loga^b = f(a^x)
\n" ); document.write( "C) log b = x= a^x
\n" ); document.write( "D) loga^b = x
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Algebra.Com's Answer #144118 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
If \"a\" is a real number and $\"a%5Ex=b\"$ , we can isolate \"x\" by applying the logarithm base 10 to both sides to get $\"log%2810%2C%28a%5Ex%29%29=log%2810%2C%28b%29%29\"$ . \r
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\n" ); document.write( "\n" ); document.write( "From there, pull down the exponent \"x\" to get $\"x%2Alog%2810%2C%28a%29%29=log%2810%2C%28b%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now divide both sides by $\"log%2810%2C%28a%29%29\"$ to get $\"x=log%2810%2C%28b%29%29%2Flog%2810%2C%28a%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Combine the terms on the right side (using the change of base formula) to get $\"x=log%28a%2C%28b%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "So after isolating \"x\", we get $\"x=log%28a%2C%28b%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Note: it turns out that $\"b%5Ey=x\"$ <====> $\"log%28b%2C%28x%29%29=y\"$ . In other words, we can convert to and from exponential and logarithmic form. \n" ); document.write( "

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