0
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Your second line is already wrong. You have to divide the logarithms, not subtract them. That said, the question is properly asked, so +1 $\endgroup$
    – Yuriy S
    Commented Feb 20, 2021 at 19:13
  • $\begingroup$ What is the basis of what you denote $\log$? $\endgroup$
    – Bernard
    Commented Feb 20, 2021 at 19:17
  • $\begingroup$ 10. As far as I concerned, base is 10 when it is not denoted. $\endgroup$
    – tahasozgen
    Commented Feb 20, 2021 at 19:20
  • $\begingroup$ 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" math.meta.stackexchange.com/questions/5020/… $\endgroup$ Commented Feb 20, 2021 at 19:39

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$
2
  • $\begingroup$ I think you know that, for example $\log_4{64}=\log_4{4^3}=3\log_44=3$ $\endgroup$ Commented Feb 20, 2021 at 19:26
  • $\begingroup$ Yes I know. Even though J.G. is the first answer, your answer is more elaborative. $\endgroup$
    – tahasozgen
    Commented Feb 20, 2021 at 19:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .