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Following is the problem: $$2\frac{\log\sqrt{27}+\log\sqrt{64}-\log\sqrt{343}}{\log144-\log49}$$

where $\log x = \log_{10}x$

My method: \begin{align*} 2\frac{\log27^{\frac{1}{2}}+\log64^{\frac{1}{2}}-\log343^{\frac{1}{2}}}{\log144-\log49} & = \frac{\log27+\log64-\log343}{\log \frac{144}{49}} \\ & = \frac{\log \frac{27\cdot 64}{343}}{\log \frac{144}{49}} \\ & = \frac{27\cdot 64}{343}\frac{49}{144} \\ & = \frac{12}{7}. \end{align*} But wolfram-alpha shows the answer as $\frac{3}{2}$

What went wrong here?

ViktorStein
  • 4,838
  • @Ramanujan The base of the logs don't matter as long as they are consistent throughout. Because the expression is basically the ratio of two logs. – Deepak Jul 31 '21 at 12:46

4 Answers4

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You REALLY need to review logarithms if you think $\frac{loga}{logb} = \frac{a}{b}$.

One of the basic rules of logarithms is $$ log_aM = log_ab \cdot log_bM $$ (The proof is rather simple and can be derived from the definition of logarithm.)

Assuming the base to be 10 when not specified, we can say that $$ \frac{log\frac{27\cdot64}{343}}{log\frac{144}{49}} = log_{\frac{144}{49}}\frac{27\cdot64}{343} = log_{\frac{144}{49}}\frac{\sqrt{144^3}}{\sqrt{49^3}} = \frac{3}{2} $$

onigari
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From this step:

$= \frac{\log \dfrac{27\cdot 64}{343}}{\log \dfrac{144}{49}}$,

you have,

$ \frac{\log \dfrac{12^3}{7^3}}{\log \dfrac{12^2 }{7^2 }} = \dfrac{3\log \dfrac {12}{7}}{2\log \dfrac {12} 7} = \dfrac 32$.

Deepak
  • 26,801
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The second to last step is incorrect: $\log a\log b\neq\frac{a}{b}$.

However, you may notice that $$\log_{\frac{144}{49}}\left(\frac{27\cdot64}{343}\right)=\frac{\log\left(\frac{27\cdot64}{343}\right)}{\log\left(\frac{144}{49}\right)}$$

We also note that

$$\frac{27\cdot64}{343}=\frac{3^34^3}{7^3}, \frac{144}{49}=\frac{3^24^2}{7^2}$$

So,

$$\left(\frac{144}{49}\right)^{\frac32}=\left(\frac{3^24^2}{7^2}\right)^{\frac32}=\frac{3^{2\cdot\frac32}4^{2\cdot\frac32}}{7^{2\cdot\frac32}}=\frac{3^34^3}{7^3}=\frac{27\cdot64}{343}$$

Therefore,

$$\frac{\log\left(\frac{27\cdot64}{343}\right)}{\log\left(\frac{144}{49}\right)}=\boxed{\frac32}$$

(Note: It doesn't matter what base your original log is, whether it's $10$ or $e$ or $2$ or any other reasonable base)

Kyan Cheung
  • 3,184
  • Can you explain from where did you get $\frac{144}{49}^\frac{3}{2}$ –  Jul 31 '21 at 12:47
  • @Sujay Note $\frac{3^34^3}{7^3}=\left(\frac{3\cdot4}{7}\right)^3$ and $\frac{3^24^2}{7^2}=\left(\frac{3\cdot4}{7}\right)^2$, hence $\left(\frac{3^24^2}{7^2}\right)^{\frac32}=\left(\frac{3\cdot4}{7}\right)^{2\cdot\frac32}=\left(\frac{3\cdot4}{7}\right)^3=\frac{3^34^3}{7^3}$. – Kyan Cheung Jul 31 '21 at 15:40
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The problem is $\dfrac{\log a}{\log b}$ is not equal to $\dfrac{a}{b}$

Instead we know $\dfrac{\log a}{\log b}=\log_b a$

Now solve this

Now $$\log_{\dfrac{144}{49}}\dfrac{27\cdot64}{343} = \log_{\dfrac{144}{49}}\frac{\sqrt{144^3}}{\sqrt{49^3}}$$

ViktorStein
  • 4,838