# Logarithm rules and how they work as algebra

I am having trouble understanding how logarithms work. If

$\log_a x = n$ that is the same as $a^n = x$

So if that is true than the log rule

$\log_b(x^n) = n \log_b x$ is the same as $a^z = x^n$ which is the same as $n(a^z) = x$

But using numbers for this it is not true. What went wrong?

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I think $logb(x^n) = n logb^x$ has error –  Babak Miraftab May 4 '12 at 18:35
Reformatted the text, but I didn't fix the errors in text because they are the source of the problem, I think –  Thomas Andrews May 4 '12 at 18:42
Also, note that you didn't define $z$ anywhere. –  Thomas Andrews May 4 '12 at 18:43
Finally, note that you mention $b$ in the second step, but you don't use it on the right hand side. –  Thomas Andrews May 4 '12 at 18:47
How did you go from $a^z = x^n$ to $n(a^z) = x$? –  TMM May 4 '12 at 20:06

Indeed, $\log_b(r)=s$ means that $b^s=r$.

So, let's look at $\log_b(x^n)$ and $n\log_b(x)$. Let $t=\log_b(x^n)$; this means that $b^t = x^n$. And let $u = \log_b(x)$. That means that $b^u = x$.

Then $n\log_b(x) = nu$. Now, what is $b^{nu}$? $$b^{nu} = b^{un} = \left(b^u\right)^n = x^n = b^t$$ (because $b^u = x)$. So, indeed, $nu=n\log_b(x)$ is the same thing as $t=\log_b(x^n)$.

Every rule of logarithms corresponds to a rule of exponentiation. Here, the rule that says that $\log_b(x^n) = n\log_b(x)$ corresponds to the rule of exponentiation that says that $$(b^r)^s = b^{rs}.$$

For another example, the rule that says that $\log_b(xy) = \log_b(x) + \log_b(y)$ corresponds to the exponentiation rule that says that $b^rb^s = b^{r+s}$. Indeed, say $r=\log_b(x)$ and $s=\log_b(y)$. Then $b^r=x$, $b^s=y$, so $xy = (b^r)(b^s) = b^{r+s}$, so $\log_b(xy) = r+s = \log_b(x)+\log_b(y)$.

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$\log_b(x^n) = n \log_b x$ is incorrect,it should be $\log_b(x^n)=n\log_b(x)$,so in case of $a^z=x^n$ $\ln(a^z)=\ln(x^n)$ ,so $z\ln a =n\ln x$ ,from here $\ln x=(z/n)\ln a$ and finally $x=\exp((z/n)\ln a)$ or $x=a^(z/n)$( i wanted to write a in power of $z/n$)

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I thought of that as a formatting error. His earlier text said $log_a^x$ when he clearly meant $\log_a x$. The real error is later, when he says $n(a^z) = x$, where $z$ is presumable $\log_b(x^n)$ –  Thomas Andrews May 4 '12 at 18:44