document.write( "Question 408682: Solve the following exponential equations using logarithms:\r
\n" ); document.write( "\n" ); document.write( "2^(x) = 3^(x+1)\r
\n" ); document.write( "\n" ); document.write( "Feeling pretty lost on this one...
\n" ); document.write( "I've tried to google for similar examples, but when I do find a similar problem it's solves using a form like this: \"xln(2)=(xln(3)+ln(3))\" I think this is a little beyond where I am now :( Looking for a more basic algebra solution! \r
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\n" ); document.write( "Thanks :)
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Algebra.Com's Answer #287822 by richard1234(7193)\"\" \"About 
You can put this solution on YOUR website!
The ln function shouldn't be too far beyond your reach, since ln is a logarithm function (except it uses the base \"e\", where e = 2.71828.... e is a very important number in calculus).\r
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\n" ); document.write( "\n" ); document.write( "The easiest way to solve this is to isolate the 1 on the exponent x+1 to obtain\r
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\n" ); document.write( "\n" ); document.write( "\"2%5Ex+=+3%2A3%5Ex\" Divide both sides by \"3%5Ex\"\r
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\n" ); document.write( "\n" ); document.write( "\"%282%5Ex%29%2F%283%5Ex%29+=+3\"\r
\n" ); document.write( "\n" ); document.write( "\"%282%2F3%29%5Ex+=+3\"\r
\n" ); document.write( "\n" ); document.write( "Take the log base 2/3 of both sides to obtain\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+log%282%2F3%2C+3%29\" = -2.709511...
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