0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What does your calculator show if you plot the range $x=50..100?$ $\endgroup$ – gammatester Dec 6 '18 at 16:36
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Piano Land Dec 6 '18 at 16:25
  • $\begingroup$ I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one. $\endgroup$ – Eclipse Sun Dec 6 '18 at 16:28
  • $\begingroup$ Perhaps just because $x=5^{8/3}$ is too large. $\endgroup$ – Eclipse Sun Dec 6 '18 at 16:38
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @KM101 Opssss...thanks I fix! $\endgroup$ – gimusi Dec 6 '18 at 16:26
  • $\begingroup$ No problem! :-) $\endgroup$ – KM101 Dec 6 '18 at 16:36
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.