# 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 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:

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 $$\frac{5}{3}$$?

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

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, $$x = \boxed{\frac{5}{3}}$$
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 $$x$$.
• Very good answer and very good intuition ! $(+1)$ May 6 '21 at 14:00
$$3=12x-17$$ $$12x=20$$ $$x=\frac {20} {12}$$