Question

logbase4(0.25)+logbase2(0.0625)

Answer 1

Really Big Hints

\(0.25 = 4^{-1}\)

\(0.0625 = 2^{-4}\)

Answer 2

please tell me the final answer

Answer 3

Please read the Code of Conduct. There should be a link on the lower left of your screen. I am constrained from simply giving answers. Show your work and let's talk.

Answer 4

\[\log_{4} 0.25+\log_{2} 0.0625 is\]

Answer 5

Retyping the problem statement does not constitute any work shown.

A logarithm IS an exponent. Those were REALLY BIG hints. Give it some thought.

Answer 6

ok

Answer 7

my answer is 5/16 is that right

Answer 8

Writing down an answer does not constitute showing work. No, 5/16 is not correct. Show my how you managed that answer.

Answer 9

\[4^{-1}+2^{-4}=1/4+1/2^{4}=1/4+1/16=5/16\]

Answer 10

Why did you just discard the logs? There is no rule like that. You must use rules.

\(\log_{4}(0.25) = \log_{4}\left(4^{-1}\right) = (-1)\cdot\log_{4}(4) = (-1)\cdot (1) = -1\)

There is no shortcut. You do the other piece.

Answer 11

ok I got that but I hv a question hoe did u get \[\log_{4}0.25=\log_{4}(4^{-1}) \]

Answer 12

\(0.25 = \dfrac{1}{4} = 4^{-1}\)

Answer 13

\[\log_{2}0.0625=\log_{2}(0.25)^{2}=2\log_{2}0.25=2\log_{2}(1/4)=2\log_{2}(4^{-1})=2\log_{2}(2^{-2})=2x-2\log_{2}^{2}=-4x1=-4 \]

Answer 14

Ran off the edge. That's fine but it's a little laborious. Since the Base is 2, I am caused to wonder if there is an integer power of 2 in that argument.

\(2^{-1} = 0.5\)
\(2^{-2} = 0.25\)
\(2^{-3} = 0.125\)
\(2^{-4} = 0.0625\) -- Aha!

Answer 15

okay

Answer 16

another question \[\log_{\sqrt{5}}0.008+\log_{2*\sqrt[3]{5?}} 1600 1s \]

Answer 17

\[\log_{\sqrt{5}} 0.008\]

Answer 18

Wow! What's the point of this question?

Anyway, I might be tempted to transform to a common base.

\(\log_{b}(a) = \dfrac{\log_{c}(a)}{\log_{c}(b)}\) for suitable a, b, and c.

Answer 19

Or not.

\(0.008 = 8/1000 = 4/500 = 2/250 = 1/125 = 5^{-3}\)

Answer 20

Do you see it?

Answer 21

okay I got that

Answer 22

Then \(5 = (\sqrt{{5}})^{2}\), that \(5^{-3} = (\sqrt{5})^{-6}\) and we're almost done.

I need to go. You tackle that other piece. Keep in mind, since it has been the nature of all the problems we have seen, that you may just want to calculate some integer exponents of the base and see how long it takes to find the argument.

Answer 23

1. \[\log_{8}128-\log_{9}\cot (\pi/3) \] is