document.write( "Question 268217: How do you solve the exponential equation in this problem?
\n" ); document.write( "3^x=243
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Algebra.Com's Answer #196627 by jsmallt9(3758) $\"\"$ $\"About$

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$\"3%5Ex+=+243\"$
\n" ); document.write( "So we are trying to find the power of 3 that is equal to 243. Well
\n" ); document.write( " $\"3%5E0+=+1\"$
\n" ); document.write( " $\"3%5E1+=+3\"$
\n" ); document.write( " $\"3%5E2+=+9\"$
\n" ); document.write( " $\"3%5E3+=+27\"$
\n" ); document.write( " $\"3%5E4+=+81\"$
\n" ); document.write( " $\"3%5E5+=+243\"$
\n" ); document.write( "So the answer is 5.

\n" ); document.write( "If the right side had been 242 or if we din't even think to see if 243 is a whole number power of 3, then we would use logarithms to solve an equation like this. Use a logarithm your calculator \"knows\", like base 10 or base e (ln):
\n" ); document.write( " $\"log%28%283%5Ex%29%29+=+log%28%28243%29%29\"$
\n" ); document.write( "Now we use a property of logarithms, $\"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\"$ , to move the coefficient out in front. (This property is the reason for using logarithms. It gives us a way to get the variable out of the exponent.):
\n" ); document.write( " $\"x%2Alog%28%283%29%29+=+log%28%28243%29%29\"$
\n" ); document.write( "Divide both sides by $\"log%28%283%29%29\"$ :
\n" ); document.write( " $\"x+=+log%28%28243%29%29%2Flog%28%283%29%29\"$
\n" ); document.write( "If you use your calculator to find both of these logarithms and then divide them, you will find that the answer is 5 (or a decimal number very, very close to 5. Remember not even your calculator know what log(243) or log(3) are exactly. It must use decimal approximations for them. And these decimal approximations will have round-off errors which may result in an answer like 5.00000001 or 4.9999999.) \n" ); document.write( "

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