document.write( "Question 419035: log(base 7)3^(x-1)=2 \n" ); document.write( "

Algebra.Com's Answer #293143 by jsmallt9(3758) $\"\"$ $\"About$

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$\"log%287%2C+%283%5E%28x-1%29%29%29=2\"$
\n" ); document.write( "Solving equations where the variable is in an exponent usually involves the use of logarithms. Logarithms are used because thay have a property, $\"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\"$ , which allows you to move an exponent out in front where we can \"get at it\" to solve for the variable.

\n" ); document.write( "Your equation already has the logarithm in it. So we go straight to using the property to move the exponent:
\n" ); document.write( " $\"%28x-1%29log%287%2C+%283%29%29=2\"$
\n" ); document.write( "To solve for x we will start by dividing both sides by that logarithm:
\n" ); document.write( " $\"x-1=%282%2Flog%287%2C+%283%29%29%29\"$
\n" ); document.write( "And last of all, add 1:
\n" ); document.write( " $\"x=%282%2Flog%287%2C+%283%29%29%29%2B1\"$
\n" ); document.write( "This is an exact expression for the solution. If you need/want a decimal approximation, them use the change of base formula for logarithms, $\"log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29\"$ , to convert the base 7 log into an expression of logs whose base your calculator \"knows\", like base 10 or base e (aka ln). Then use your calculator to find the two logarithms and simplify. \n" ); document.write( "

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