Logarithm Problem-Simplification of Long Problems

Question

I can't figure out how to solve this problem, so it would be awesome if someone could tell me the technique I need for problems of this sort (as applied to this problem). The answer is 23, but I'm not sure how to get to the answer. I think the first part equals 11, but I'm probably totally off.

$$5^{\log_5 22-\log_5 2} +3 \; \log_2 2^4$$

Answer 1

Note that

$$5^{\log_5 22-\log_5 2} +3 \; \log_2 2^4=\frac{5^{\log_5 22}}{5^{\log_5 2}} +12 \; \log_2 2=\frac{22}{2}+12=23$$

we have used that

• $a^{b-c}=\frac{a^b}{a^c}$
• $a^{\log_ab}=b$
• $\log_ab^c=c\log_a b$

Answer 2

use that $$\log_5 22-\log_5 2=\log_5\left(\frac{22}{2}\right)=\log_5 11$$ and $$\log_2 2^4=4\log_2 2=4$$ and $$5^{\log_5 11}=11$$

Answer 3

Some generally useful facts are that for any positive base $b,$ $\log_b x + \log_b y = \log_b(xy)$ and $\log_b x - \log_b y = \log_b\left(\frac xy\right).$ You can use $\log_b x - \log_b y$ to quickly simplify $\log_5 22-\log_5 2.$ (Another answer has already used this rule this way.)

For the rest of it, remember that $x \mapsto \log_b(x)$ and $y \mapsto b^y$ are inverse functions of each other: $$ b^{\log_b x} = x \qquad\text{and}\qquad log_b(b^y) = y . $$

Using these facts you can quickly deal with $5^{\log_5 11}$ and $\log_2 2^4.$