Calculating $\log_7 125$ So the problem asks to calculate $\log_7 125$.
It's multiple choice and the options are


*

*$2.48$

*$4.75$

*$1.77$

*$2.09$


Given that $7^2 = 49$ and $7^3 = 343$, the answer must be either option 1 or 4, not 2 or 3.
So now what.
I remembered there's a way to translate bases like so:
$$
\log_a x = (\log_a b)(\log_b x)
$$
which translates to
$$
\log_7 125 = (\log_7 5)(\log_5 125)
$$
which is 
$$
3\log_7 5
$$
But then what?
I didn't know so I took an educated guess and went which option 1, which was right.
But for next time, what should I do?
What is the general strategy for solving problems like this when the base and the number have no obvious relationship?
 A: We estimate $$7^{2.48}\approx 7^{2.5}=7^2\sqrt{7}\approx 49\times 2.5\approx 125$$
On the other hand, $$7^{2.09}\approx 7^2\sqrt[10]{7}\approx 49\times 1.2\approx 60$$
A: Apparently, they were counting on you to make the following mistake, by confusing $\log_7$ with the natural logarithm $\ln$, which has the property that $\ln(1+x)\simeq x$ for small values of x :
$$\log_75=\log_7(7-2)=\log_77\left(1-\frac27\right)=\log_77+\log_7\left(1-\frac27\right)\simeq1+\left(-\frac27\right)=\frac57$$
$$3\log_75\simeq3\cdot\frac57=\frac{15}7=2\frac17\simeq2.09$$
Likewise, the $4.75$ value probably comes from a similarly wrong approach, but with $+$ instead of $-$ in the above formula.
A: A quick estimate :
$$\log_7 125\approx \frac{\log(2^7)}{\log(2^3)}\approx \frac 73$$
Something a little more precise (since $125=50\cdot 2.5$) :
$$\log_7 125\approx \frac{\log(49\cdot \sqrt{7})}{\log(7)}\approx \frac{\log(7^{2.5})}{\log(7)}\approx 2.5$$
A: A general strategy that can be done by hand computation to compute $\log_b(a)$.
We use the fact that $\log_b(a)=n+\log_b(\frac{a}{b^n})$ and $\log_b(a^2)=2\log_b(a)$
Find the largest integer $n$ such that $b^n\le a$. 
Let $a_1=(\frac{a}{b^n})^2$ and define :


*

*if $a_n<b$ then $b_n=0$ and $a_{n+1}=(a_n)^2$

*if $a_n\ge b$ then $b_n=1$ and $a_{n+1}=(\frac{a_n}{b})^2$


Then $\log_b(a)=n+\sum_{i=1}^\infty b_i 2^{-i}$
Of course by hand, you only need to keep 2 or 3 significant digits each time to get a good approximation.
Here $\log_7(125)$, you found $n=2$


*

*$a_1=(\frac{125}{49})^2$. By hand $125\div49\approx 2.55$, Hence $a_1\approx(2.55^2)\approx 6.5$ : $a_1<7$ $b_1=0$ (so the result is less than $2.5$ but it must be close as $6.5$ is close to $7$)

*$a_2\approx(6.5)^2\approx 42$ : $a_2>7$ $b_2=1$ (so the result is in $[2.25,2.5)$)

*$a_3\approx (\frac{42}{7})^2=6^2=36$, $a_3>7$ $b_3=1$ (so the result is in $[2.375,2.5]$) 

*and so one…

