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.

algebra-precalculuslogarithms

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 $\frac{5}{3}$?

Best Answer

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, $x = \boxed{\frac{5}{3}}$

Related Solutions

Solve $3+e^{2t}=7e^{2t}$

Just note that $$t_{\text{yours}}=\frac{1}{2}\ln(\frac{1}{2})=\color{blue}{\frac{1}{2}\ln(2^{-1}) = \frac{1}{2}(-1)\ln(2)} = -\frac{1}{2}\ln(2) = t_{\text{book's}}$$

Using definition of logs, solve $5\log_7(n)=10$

What you did wrong is $n=7^{10-5}$ does not follow from $7^{10}=n^5$.

To fix your calculation (as WA Don pointed out in a comment), $$ 7^2=(7^{10})^{1/5}=(n^5)^{1/5}=n $$

Alternatively,

\begin{align} 5\log_7(n)&=10\\ \log_7(n)&=2&\textrm{(dividing by $5$)}\\ n&=7^2=49&\textrm{($\log_ax=y$ iff $x=a^y$)} \end{align}

For your meta question:

\begin{align}
5\log_7(n)&=10\\
\log_7(n)&=2&\textrm{(dividing by $5$)}\\
n&=7^2=49&\textrm{($\log_ax=y$ iff $x=a^y$)}
\end{align}

Best Answer

Related Solutions

Solve $3+e^{2t}=7e^{2t}$

Using definition of logs, solve $5\log_7(n)=10$

Related Question