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

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.

• What does your calculator show if you plot the range $x=50..100?$ – gammatester Dec 6 '18 at 16:36

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 $$y=0,\quad y=\frac{-1+\sqrt{33}}{2},\quad y=\frac{-1-\sqrt{33}}{2}$$

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.

• But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1. – Piano Land Dec 6 '18 at 16:25
• I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one. – Eclipse Sun Dec 6 '18 at 16:28
• Perhaps just because $x=5^{8/3}$ is too large. – Eclipse Sun Dec 6 '18 at 16:38

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.

• @KM101 Opssss...thanks I fix! – gimusi Dec 6 '18 at 16:26
• No problem! :-) – KM101 Dec 6 '18 at 16:36

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.

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 $$\log_5 x$$.

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

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