document.write( "Question 382174: logx(2.5)=2.6 please help me solve. \n" ); document.write( "

Algebra.Com's Answer #270949 by jsmallt9(3758) $\"\"$ $\"About$

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$\"log%28x%2C+%282.5%29%29=2.6\"$
\n" ); document.write( "One way to solve this start with rewriting the equation in exponential form. In general $\"log%28a%2C+%28p%29%29+=+q\"$ is equivalent to $\"p+=+a%5Eq\"$ . Using this pattern on your equation we get:
\n" ); document.write( " $\"2.5+=+x%5E2.6\"$

\n" ); document.write( "From here we are trying to get an equation of the form:
\n" ); document.write( "x = some-expression
\n" ); document.write( "\"x\" has an exponent of 1. (Any \"invisible\" exponent is always a 1.) The equation we have has x with an exponent of 2.6. So somehow we need to find a way to change the exponent of 2.6 into a 1. To do this we will combine the following ideas:

When raising a power to a power, the rule for exponents is to multiply the exponents.
Multiplying reciprocals always results in a 1.

\n" ); document.write( "So to change the exponent on x in the desired way, we are going to raise both sides of the equation to the reciprocal of 2.6 power. The reciprocal of 2.6 is $\"1%2F2.6+=+1%2F%2826%2F10%29+=+10%2F26+=+5%2F13\"$
\n" ); document.write( " $\"%282.5%29%5E%285%2F13%29+=+%28x%5E2.6%29%5E%285%2F13%29\"$
\n" ); document.write( "which simplifies to:
\n" ); document.write( " $\"%282.5%29%5E%285%2F13%29+=+x\"$
\n" ); document.write( "This is an exact expression for the solution to your equation. If you want a decimal approximation then you can use your calculator on the left side. (If your calculator has buttons for parentheses, then you can type in 2.5^(5/13)
\n" ); document.write( "If your calculator does not have buttons for the parentheses. Then divide 5 by 13 and record the answer. Then type 2.5^ followed by the decimal answer to 5/13.) \n" ); document.write( "

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