algebra-precalculuslogarithms
I am to solve $5\log_7(n)=10$. I arrived at $n=7^5$ while the solution in my textbook says it's $49$
My working:
$$5\log_7(n)=10$$
$$\log_7(n^5)=10$$
$$7^{10}=n^5$$
$$n=7^{10-5}$$
$$n=7^5$$
Where did I go wrong and how can I arrive at 49?
(Meta question, is there a way that I can add comments next to each line of my working to say what I was thinking? I tried using '#' next to each line but that buckled the markup)
The root is $x = -4$. So you have to take $-4$ instead of $4$
Hence you have $2x^2-2x-3$
Let's go step by step.
We start with the fact that $8=2^3$. That's equivalent to $2=8^{1/3}$ from which we immediately deduce that $$\log _{8} 2=\frac{1}{3}$$
Now, of course, we really wanted $\log_84$. But, as $4=2^2$ we have $$\log_84=\log_82^2=2\log_82=\frac 23$$ It follows that $$\boxed{6\log_84=6\times \frac 23=4}$$
Best Answer
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: