# Given $a = \log_{\sqrt{6}}{\sqrt[3]{4}}$ find $\log_{\sqrt{3}}{\sqrt[3]{2}}$

I have attempted to solve this and got the result of $$\frac{2a}{4-3a}$$ Whereas the correct answer: $$\frac{10a}{12-15a}$$ I have attached my calculations. Could you tell me where they are inaccurate?

$$a=\log_{\sqrt{6}}\sqrt[3]{4}$$ $$\log_{\sqrt{3}}\sqrt[3]{2}=\frac{1}{3}\log_{\sqrt{3}}2=\frac{1}{3}\log_{\sqrt{3}}4^{1/2}=\frac{1}{6}\log_{\sqrt{3}}4=\frac{1}{2}\cdot \frac{1}{3}\log_{\sqrt{3}}4$$ $$=\frac{1}{2}\log_{\sqrt{3}}\sqrt[3]{4}=\frac{1}{2}\cdot \frac{\log_{\sqrt{6}}\sqrt[3]{4}}{\log_{\sqrt{6}}\left(\frac{\sqrt{6}}{\sqrt{2}}\right)}=\frac{1}{2}\cdot \frac{a}{1-\log_{\sqrt{6}}\sqrt{2}}$$ $$=\frac{a}{2(1-\frac{1}{2}\log_{\sqrt{6}}2)}=\frac{a}{2(1-\frac{1}{4}\log_{\sqrt{6}}4)}=\frac{\frac{1}{3}a}{\frac{2}{3}-\frac{2}{12}\log_{\sqrt{6}}4}$$ $$=\frac{\frac{1}{3}a}{\frac{2}{3}-\frac{2}{4}\cdot \frac{1}{3}\cdot \log_{\sqrt{6}} 4}=\frac{\frac{1}{3}a}{\frac{2}{3}-\frac{1}{2}a}=\frac{\frac{1}{3}a}{\frac{4-3a}{6}}=\frac{2a}{4-3a}$$

Hence, the final answer is: $$\bbox[5px,border:2px solid #C0A000]{\log_{\sqrt{3}}\sqrt[3]{2}=\frac{2a}{4-3a}}$$

• That's very hard to read – John Doe Apr 27 '17 at 20:41
• I did my best to make it as legible as possible. Which part can't you read? – ILoveChess Apr 27 '17 at 20:42
• Your answer seems completely correct, perhaps the answer given is not correct? Edit: checked on a calculator and your answer was right. – John Doe Apr 27 '17 at 20:46
• Is my edit correct? If so, you may delete the picture. – projectilemotion Apr 27 '17 at 20:51
• @projectilemotion Yes, thank you very much! – ILoveChess Apr 27 '17 at 20:53

I followed your method, and every step was correct. Checking the answer in a calculator I get that $$\log_{\sqrt{3}}{\sqrt[3]{2}} = 0.42061...$$
and $$\frac{2a}{4-3a} = 0.42061...=\log_{\sqrt{3}}{\sqrt[3]{2}}\quad\text{as required}$$
whereas $$\frac{10a}{12-15a} = 1.20997...\quad\text{which is clearly incorrect}$$