Question

1/4 log 2 16 + 1/2 log 2 49 = log 2 x
What is the solution to this?

Answer 1

Please help!

Answer 2

@KayJanae I'll help you. Do you know the basic properties of logarithms?

Answer 3

No I dont

Answer 4

okay, the following are important
\[1)\log_x y=z \longrightarrow y=x^z\]
\[2) a\log b=\log b^a\]
\[3) \log x+\log y=\log xy\]
\[4) \log x-\log y=\log\frac xy\]

Answer 5

Alright, and do they have a certain name?

Answer 6

Ignore that question lol where do we go from here?

Answer 7

Yeah, let me pull a website, where will you get the names also

Answer 8

Ok, thanks

Answer 9

do check this
http://en.wikipedia.org/wiki/Logarithm
let's work on the problem

Answer 10

ok

Answer 11

We have
\[\frac 14 \log_2 16 +\frac 1 2 \log_2 49 = \log_2 x\]
one more thing, all those properties only if the base is same

Answer 12

Ok, the base is 2 right?

Answer 13

yeah, here all have same, so we could use the properties
tell me what's this ?
\[\frac 14 \log_2 16\]
use one of the properties to simplify

Answer 14

Um, how do i do that?

Answer 15

\[16^{1/4} * 49^{1/2} =x\]

Answer 16

That isnt right. It isnt one of the choices

Answer 17

\[16^{1/4} * 49^{1/2} = 2*7 =14\]

Answer 18

Ohh ok thank you, could you do another one for me?

Answer 19

log 3 0.1 + 2log 3 x = log 3 2 + log 3 5

Answer 20

The principle is the same, just as ash2326 showed you:
\[0.1 * x ^{2} = 2 * 5\]
that's what your above expression reduces to, remember when you add logs, you multiply the real numbers, and when you subtract logs, you divide the real numbers. can you try solve this one

Answer 21

\[0.1x ^{2} = 10\]
\[x ^{2} = 10/0.1 = 100\]
\[x=10\]

Answer 22

ok.. i think i understand, so the answer to my question is 10?

Answer 23

yh