$x^{\log_5 3}=-6$,Solve for x

Question

This question feels banal to ask, but I feel lost. Here is what I did.

$x^{\log_5 3}=-6$

$\ln(x^{\log_5 3})=\ln(-6)=\ln(6)+i\pi$

$\log_53* \ln(x)=\ln(6)+i\pi$

$\ln(x)=\frac{\ln(6)+i\pi}{\log_53}=(\ln(6)+i\pi)\log_35$

$x=e^{\ln(6)+i\pi)\log_35}$

This seems right to me, but when I verify with wolfram it shows that I am wrong.

https://www.wolframalpha.com/input/?i=(e%5E(ln(-6)log(3,5)))%5E(log(5,3))

Answer 1

Update: I originally thought the input text to Wolfram was causing a double exponentiation. I later realized this is not true, with the issue instead being due to how the use of brackets caused it to interpret the calculations. I have now corrected my answer below.

$ $

As commented by several people, your solution is basically correct. The issue with your use of Wolfram is that your input text is causing it to interpret it differently than you intended due to your use of brackets. You asked it to find "(e^(ln(-6)log(3,5)))^(log(5,3))". Using your specific instructions, Wolfram is calculating, as it says in the exact result section:

$$\left(5^{\left(\left(log(6) + i π\right)/log\left(3\right)\right)}\right)^{(log(3)/log(5))} \tag{1}\label{eq1}$$

i.e., it's using a base of $5$ instead of the intended $6$. This is due to your use of brackets and a power separate from the original expression, causing Wolfram to reinterpret the part inside the large brackets on their own, where the radius in polar coordinates is $5$, before then applying the final power. However, as the "Polar coordinates" part shows, the radius of the overall expression is still $6$ (as the exponent outside of the main brackets of $\log\left(3\right)/\log\left(5\right)$ causes it to be changed), but with a $\theta$ of about $-65.7382°$ instead of the $\pi$ radians, i.e., $180°$, which you wanted. As Mark Viola says in his answer, the logarithm function with complex variables is multi-valued. Thus, you need to be careful how you use it, including how you express it to Wolfram to calculate for you.

The better expression to use with Wolfram is something like "e^((ln(-6)log(3,5))(log(5,3)))" (note: I'm not using a direct link with the equation in Wolfram as the URL is not being handled properly), i.e., remove your second "^" and adjust your braces. This puts all of your work together within the same brackets, in particular it has the multiplication of the exponents together, so Wolfram interprets it correctly for you, i.e., it gives you back a result of $-6$ as you expected.

Answer 2

The definition of $z^C$, where $z\in\mathbb{C}$ and $C\in \mathbb{C}$, is the multivalued number given by

$$z^C=e^{C\left(\log(|z|+i\arg(z)) \right)}$$

where $\arg(z)$ is multivalued and $\log(|z|)$ is the logarithm from real analysis.

Now, suppose $C=\log_5(3)$ and $z=x\in\mathbb{C}$, we have

$$x^{\log_5(3)}=|x|^{\log_5(3)}e^{i\log_5(3)\arg(x)}$$

Setting $x^{\log_5(3)}=-6=6e^{i(2n+1)\pi}$ we find that the magnitude of $x$ is

$$|x|=6^{\log(5)/\log(3)}$$

and the argument of $x$ is

$$\arg(x)=(2n+1)\pi\frac{\log(5)}{\log(3)}$$

Therefore, the solution for $x$ is

$$x=e^{\log(5)\log(6)/\log(3)}e^{i(2n+1)\pi\frac{\log(5)}{\log(3)}}$$