1
$\begingroup$

I attempted the question in the title:

Rewrite $5^{12x-17}=125$ as a logarithm. Then apply the change of base formula to solve for x using the common log. Round to the nearest thousandth.

I arrived at $x=\frac{14}{12}$ whereas my textbook says the solution is actually this:

enter image description here

My working:

$$5^{12x-17}=125$$ $$\log_5(125)=12x-17$$ $$\frac{\ln(125)}{\ln(5)}=12x-17$$ $$3=12x-17$$ $$12x=14$$ $$x=\frac{14}{12}$$

Where did I go wrong and how can I arrive at $\frac{5}{3}$?

$\endgroup$
1
  • 2
    $\begingroup$ Wait, how can you go from $3=12x-17$ to $12x=14$? (Yes, an arithmetic error.) $\endgroup$ May 5 '21 at 12:42
4
$\begingroup$

In your 4th step, you said $3 = 12x - 17$ then in your 5th step, you said $12x = 14,$ when it's actually $12x = 17 + 3 = 20.$ So, $x = \boxed{\frac{5}{3}}$

$\endgroup$
2
  • $\begingroup$ This is why we lay out all of our steps I guess! I just assumed my error was further up. Thanks for spotting! $\endgroup$
    – Doug Fir
    May 5 '21 at 12:46
  • $\begingroup$ @DougFir no problem :) also nice solution! $\endgroup$ May 5 '21 at 12:46
3
$\begingroup$

Alternatively, from $5^{12x-17} = 5^3$ you can directly conclude that $12x-17 = 3 \implies x = \frac{5}{3}$ as the exponential function $a^x$ is injective (one-to-one). It is injective as its inverse exists, being the function $\frac{\ln x}{\ln a}$, which you can find by making $x$ the subject in $y = a^x$.

Note that this does not hold in general. For instance, $x^2 = 4 \implies x = 2, -2$ as when drawing the line $y = 4$, it intersects the curve at two points, hence two solutions. Since $12x-17$ can be any real number, you need this property of $1$ y-value $\implies$ $1$ x-value to hold for all real $x$.

$\endgroup$
1
  • 1
    $\begingroup$ Very good answer and very good intuition ! $(+1)$ $\endgroup$
    – Etemon
    May 6 '21 at 14:00
2
$\begingroup$

$$3=12x-17$$ $$12x=20$$ $$x=\frac {20} {12}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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