Mathematics
OpenStudy (callisto):

Uncategorised question #2 Let $\large y=\log_{1400}\sqrt{2} + \log_{1400}\sqrt[3]{5} +\log_{1400} \sqrt[6]{7}$, find the value of $$y$$. Note: No calculator + I don't have the answer to this question

OpenStudy (anonymous):

$\large y=\log_{1400}\sqrt{2} + \log_{1400}\sqrt[3]{5} +\log_{1400} \sqrt[6]{7}$

OpenStudy (callisto):

:S ^ Right form I should practise latex :|

OpenStudy (experimentx):
OpenStudy (callisto):

You may post your solution here :|

OpenStudy (experimentx):

1/6 ??

OpenStudy (callisto):

I told you I don't have the solution :| From calculator, yes.

OpenStudy (accessdenied):

1/6 is what I get as well. Basically, I just used logarithm properties to combine it into one logarithm, and then changed the radicals into rational exponents to get 'common denominators' for the exponents. Then some more logarithm properties. D:

OpenStudy (experimentx):

haven't calculated yet ... used the same idea as AccessDenied described ..

OpenStudy (callisto):

Let me try~

OpenStudy (anonymous):

$\large y=\log_{1400}\sqrt{2} + \log_{1400}\sqrt[3]{5} +\log_{1400} \sqrt[6]{7}$$= \huge \log_{2^3.5^2.7} 2^\frac12 5^\frac 13 7^\frac 16$ $= \huge \log_{(2^\frac36 5^\frac 26 7^\frac 16)^6} 2^\frac36 5^\frac 26 7^\frac 16=\frac 16$ $\large \log_{a^b} a^d = \frac db$

OpenStudy (callisto):

@AccessDenied Is your solution like FFM's ?

OpenStudy (accessdenied):

My way requires more steps than his, since I didn't even think about the formula he used. lol Sorry, I'll get the LaTeX of what I did in a sec, gotta find my mistake somewhere. D:

OpenStudy (accessdenied):

$\large{ y = \log_{1400} (\sqrt{2} + \log_{1400} (\sqrt[3]{5}) + \log_{1400} (\sqrt[6]{7}) \\ y = \log_{1400} (\sqrt{2} \times \sqrt[3]{5} \times \sqrt[6]{7}) \\ y = \log_{1400} (2^{1/2} \times 5^{1/3} \times 7^{1/6} \\ y = \log_{1400} (2^{3/6} \times 5^{2/6} \times 7^{1/6} \\ y = \log_{1400} (2^3 \times 5^2 \times 7)^{1/6} \\ y = \frac{1}{6} \log_{1400} (2^3 \times 5^2 \times 7) }$ The prime factorization of 1400 turns out to be $$2^3 \times 5^2 \times 7$$, so the base is the same as the argument of the logarithm. $$log_{b} b = 1$$ Then, we just multiply by the 1/6 remaining on the outside and we get 1/6.

OpenStudy (mertsj):

$\log_{1400}2^{\frac{3}{6}}+\log_{1400}5^{\frac{2}{6}} +\log_{1400}7^{\frac{1}{6}} =$ $\frac{3}{6}\log_{1400}2+\frac{2}{6}\log_{1400}5+\frac{1}{6}\log_{1400}7=$ $\frac{1}{6}(\log_{1400}8+\log_{1400}25+\log_{1400}7)=$ $\frac{1}{6}\log_{1400}1400=\frac{1}{6}$

OpenStudy (anonymous):

OpenStudy (callisto):

Wow!!!! I'm stupid :| Thanks all!!!!!!!!!

OpenStudy (anonymous):

You are not stupid, this was harder than the previous problem which I failed :(

OpenStudy (accessdenied):

You're welcome. :D

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