Solving the Logarithmic Equations

Question

$$5(\log_x y+\log_y x)=26 \tag1$$ $$xy=64 \tag2$$

From $(1)$, we have $$5\left(\log_x y + \frac{1}{(\log_x y)}\right)=26 \tag3$$ For the next step however, my book shows the following and proceeds to solving it: $$(\log_x y-5)\left(\log_y x-\frac{1}{5}\right)=0 \tag4$$

I tried working backwards by expanding $(4)$ and got this:

\begin{align} (\log_x y-5)\left(\log_y x-\frac{1}{5}\right)\ &=(\log_x y)(\log_y x)-\frac{1}{5}\log_x y-5\log_y x+1\\ &=\frac{\log y}{\log x}\frac{\log x}{\log y}-\frac{1}{5}\log_x y-5\log_y x+1\\ &=1-\frac{1}{5}\log_x y-5\log_y x+1\\ &=2-\frac{1}{5}\log_x y-5\log_y x\tag5\\ \end{align}

How are $(1)$ and $(5)$ related?

Answer 1

Using natural logarithms, we have

$$\frac{\log (x)}{\log (y)}+\frac{\log (y)}{\log (x)}=\frac {26}5\tag 1$$ $$\log(x)+\log(y)=\log(64)\implies \log(y)=\log(64)-\log(x) \tag2$$ Now, mka $t=\log(x)$ and plug in $(1)$ $$\frac t {\log(64)-t}+\frac {\log(64)-t}t=\frac {26}5\tag 3$$ which is just a quadratic. Solve it for $t$ and then $x=e^t$ and $y=\frac {64}x$

Answer 2

You can always view $(3)$ as a quadratic function in $t=\log_xy$ after multiplying both sides by $t$. The solutions that you get are $t=5$ and $t=\frac{1}{5}$ which then allow for the factorization of $(4)$.

Answer 3

Let set $z=\log_x(y)$ then $(3)$ is just $5(z+\frac 1z)=26\iff 5z^2-26z+5=(5z-1)(z-5)=0$

Which is $(4)$

Then use $(2)$ to get $\ln(x)=\dfrac{\ln(64)}{z+1}$