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Thread: stuck with a logarithm

  1. #1
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    stuck with a logarithm

    Hi;
    Solve for x: 4log(x - 6) = 11

    so far I'm hare log(x - 6)^4 = 11 (not sure where to go from here).

    Thanks.
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  2. #2
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    Re: stuck with a logarithm

    Apply the function inverse to log to both sides.
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  3. #3
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    Re: stuck with a logarithm

    Yeah I got it thanks.
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  4. #4
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    Re: stuck with a logarithm

    The inverse of $\displaystyle log_a(x)$ is $\displaystyle a^x$- that is if $\displaystyle y= log_a(x)$ then $\displaystyle x= a^y$. Your problem is a little ambiguous because you do not give a base for the logarithm. Normally, that would imply the "common logarithm", base 10. If that is the base here, then $\displaystyle (x- 6)^4= 10^{11}$. That is probably correct here but in higher level mathematics it is common to use "log" rather than "ln" to mean the natural logarithm. If that is the case, then $\displaystyle (x- 6)^4= e^{11}$.
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