Find the solution to $x^{(\log_5 x^2 + \log _5 x-12)}=\frac{1}{x^4}$

Question

Find the solution to $x^{(\log_5 x^2 + \log _5 x-12)}=\frac{1}{x^4}$

I equated their exponents,

That gave me $\log_5 x = \frac{8}{3}$

But the answer given in my book is $1$.

Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.

When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?

Any help would be appreciated.

Answer 1

You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base: $$ (\log_5(x^2)+\log_5x-12)\log_5x=-4\log_5x $$ that becomes, setting $y=\log_5x$, $$ (3y-8)y=0 $$ so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.

On the other hand, if the first term is $(\log_5x)^2$, rather than $\log_5(x^2)$, the equation would become $$ (y^2+y-8)y=0 $$ with solutions $$ y=0,\quad y=\frac{-1+\sqrt{33}}{2},\quad y=\frac{-1-\sqrt{33}}{2} $$

Answer 2

I suppose you were using $$x^{(\log_5 x^2 + \log _5 x-12)}=\frac{1}{x^4}\implies \log_5 x^2 + \log _5 x-12 =-4,$$ but it is only true when $x>0$ and $x\ne 1$.

Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $x\ne 1$. That's why you missed a solution.

Answer 3

If the book claims, that $1$ is the only real solution, it is wrong. Your value $$x = 5^{8/3} = e^{\frac{8}{3} \ln 5} \approx 73.1$$ is indeed a solution of the equation.

Answer 4

What you did (probably) was the following:

$$x^{\log_5 x^2+\log_5 x-12} = \frac{1}{x^4} = x^{-4} \implies \log_5 x^2+\log_5 x-12 = -4$$

By doing so, you removed the possibility of $x = 1$.

As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $\log_5 x^2$ and $\log_5 x$.

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Your other solution is valid. Let $x = 5^{\frac{8}{3}}$.

$$\log_5 \big(5^{\frac{8}{3}}\big)^2+\log_5 5^{\frac{8}{3}}-12 = \log_5 \big(5^{\frac{8}{3}}\big)^3-12 = \log_5 5^8-12 = 8-12 = -4$$

On both sides, you get

$$\big(5^{\frac{8}{3}}\big)^{-4}$$

Answer 5

We need

$$\log_5 x^2 + \log _5 x-12=-4 \iff \log_5 (x^3)=8$$

that is

$$x=5^\frac83$$

the other solution $x=1$ is obtained by inspection from the original equation.