Question

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

Answer 1

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

Answer 2

:S ^ Right form
I should practise latex :|

Answer 3

Jeez http://www.wolframalpha.com/input/?i=2^3*5^2*7

Answer 4

You may post your solution here :|

Answer 5

1/6 ??

Answer 6

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

Answer 7

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:

Answer 8

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

Answer 9

Let me try~

Answer 10

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

\[\large \log_{a^b} a^d = \frac db \]

Answer 11

@AccessDenied Is your solution like FFM's ?

Answer 12

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:

Answer 13

\[ \large{
y = \log_{1400} (\sqrt{2} + \log_{1400} (\sqrt[3]{5}) + \log_{1400} (\sqrt[6]{7}) \\
y = \log_{1400} (\sqrt{2} \times \sqrt[3]{5} \times \sqrt[6]{7}) \\
y = \log_{1400} (2^{1/2} \times 5^{1/3} \times 7^{1/6} \\
y = \log_{1400} (2^{3/6} \times 5^{2/6} \times 7^{1/6} \\
y = \log_{1400} (2^3 \times 5^2 \times 7)^{1/6} \\
y = \frac{1}{6} \log_{1400} (2^3 \times 5^2 \times 7) } \]

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.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

You're welcome. :D