A logarithmic Calculation

Question

I have encountered a logarithmic question that I could not solve. Here is the question:

$$\dfrac {\log 64}{\log4}+\dfrac {\log 8}{\log 2}$$

A)1 B)2 C)3 D)5 E)6

Rules of Logarithm:

$$\log_a\left(\frac xy\right)=\log_ax-\log_ay$$

$$\log_ax^n=n\log_ax$$

Here is my calculation:

$$Q=\dfrac {\log 2^6}{\log 2^2}+\dfrac {\log 2^3}{\log 2}$$

$$Q=6\log 2-2\log 2+3\log 2-\log 2$$

$$Q=6\log 2$$ What is the error in my calculation? Thanks in advance.

Your second line is already wrong. You have to divide the logarithms, not subtract them. That said, the question is properly asked, so +1 — Yuriy S, Feb 20 '21 at 19:13
I edited your question using mathjax. You may want to learn it simple from here. "MathJax basic tutorial and quick reference - Mathematics Meta Stack Exchange" https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — lone student, Feb 20 '21 at 19:39

score 2 · Answer 1 · answered Feb 20 '21 at 19:12

2

You've confused $\log x/\log y$ with $\log(x/y)$. Use the second rule twice, with $a=4$ ($a=2$) in the first (second) fraction.

answered Feb 20 '21 at 19:12

J.G.

score 1 · Accepted Answer · answered Feb 20 '21 at 19:20

1

Your mistake was stated in the alternative answer

$$\log_yx=\frac {\log x}{\log y}≠\log \frac xy =\log x-\log y$$

Just use

$$\frac {\log x}{\log y} =\log _yx$$

We have

$$\frac {\log 64}{\log 4}=\log_4{64}=3$$

$$\frac {\log 8}{\log 2}=\log_28=3$$

answered Feb 20 '21 at 19:20

lone student

I think you know that, for example $\log_4{64}=\log_4{4^3}=3\log_44=3$ – lone student Feb 20 '21 at 19:26
Yes I know. Even though J.G. is the first answer, your answer is more elaborative. – tahasozgen Feb 20 '21 at 19:30

2 Answers2