Question

help me evaluate these 3 log problems?

Answer 1

\[\log_\frac{ 2 }{ ?3 }\frac{ 27 }{ 8 }\]

Answer 2

\[\log_x \frac{ 1 }{ 9 }\]

Answer 3

\[\log_4x=-\frac{ 1 }{ 2 }\]
*calculators are not allowed*

Answer 4

the second question is also equal to -1/2

Answer 5

First question?

Answer 6

\[(\log_a x = y) \rightarrow (a^y = x)\] Try and use this to solve them.

Answer 7

Logarithms are basically asking, to what power does "a" need to be raised to in order for you to get "x," thus the conversion equation above.

Answer 8

I don't understand how to do that for the first question

Answer 9

wait I think got it

Answer 10

What's the answer for the second question? Just to make sure you get the concept?

Answer 11

\[\frac{ 2 }{ 3 }^x=\frac{ 27 }{ 8 }\]

Answer 12

would that be the start to number one?

Answer 13

I can't really see what number one is, it's a bit bugged. Try reinputing it?

Answer 14

\[\log_\frac{ 2 }{ 3 }\frac{ 27 }{ 8 }\]

Answer 15

Ah. Yes, you're right.

Answer 16

okay, so I get that concept just fine :) the problem is always the next part

Answer 17

Will this help you? \[\frac{ a }{ b }^{-1} = \frac{ b }{ a }\]

Answer 18

so in this case 2/1?

Answer 19

\[\frac{ a }{ b }^{-x} = \frac{ b^x }{ a^x }\]

Answer 20

-2/1

Answer 21

Ok. So we have \[\frac{ 2 }{ 3 }^x = \frac{ 27 }{ 8 }\]

We have to find one power that both 2 and 3 can be raised to in order to get 27 and 8.

\[3^3 = 27, 2^3 = 8\]

Now if you want to flip it, you merely have to add a negative sign to the exponent.

Answer 22

so make them both ^-3?

Answer 23

\[\frac{ 2 }{ 3 }^3 = \frac{ 8 }{ 27 }\] So now you just need to get \[\frac{ 27 }{ 8 }\] which can be done by changing the exponent to a negative.

Answer 24

wait so the answer is -3?

Answer 25

Yes \[\frac{ 2 }{ 3 }^ {-3} = \frac{ 27 }{ }\]

Answer 26

the 8 is missing lol. o_O but yeah.

Answer 27

ah okay! So for a more simple problem such as log_8 4 I know the answer is 2/3 but I'm not sure where I got the answer is it because 2^3 is 8 so then you make the answer 2-3?

Answer 28

2/3*

Answer 29

\[8^\frac{ 1 }{ 3 } = \sqrt[3]{8} = 2\]

Answer 30

okay I see! So is number 2 any different?

Answer 31

Different from what? o_O So \[8^\frac{ 2 }{ 3 }= \sqrt[3]{8} * \sqrt[3]{8} = 2*2 = 4\]

Answer 32

I meant problem 2 :D

Answer 33

Well no it's not, other than the negative. This may help

\[x^{-\frac{ 1 }{ y }} = \frac{ 1 }{ \sqrt[y]{x} }\]and \[x^\frac{ 1 }{ y } = \sqrt[y]{x}\]

Answer 34

So the numerator works like an exponent, but the denominator works as the "y" root of x. Does that make sense?

Answer 35

Problem 2 is basically \[\frac{ 1 }{ 9 } = x^{-\frac{ 1 }{ 2 }}\]

Answer 36

yes it does. so is it set up as:
\[\sqrt[9]{x}=\frac{ 1 }{ \sqrt{x} }\]

Answer 37

woah woah where did \[\sqrt[9]{x} \] come from?

Answer 38

Now \[\sqrt[9]{x} = x^\frac{ 1 }{ 9 }\] but \[\sqrt[9]{x} \neq \frac{ 1 }{ 9 }\]

Answer 39

so what do I do with x in this problem?

Answer 40

\[\frac{ 1 }{ 9 } = \frac{ 1 }{ \sqrt{x} }\]

Answer 41

so is the final answer 3?

Answer 42

:P Check again.

Answer 43

That's a square root, not a square, hehe.

Answer 44

so it would be 81 :D

Answer 45

Yeah haha.

Answer 46

.-. I'm so tired... I can't math... but yet I need to do 50 more problems...

Answer 47

hehe same. good luck!

Answer 48

thank you, I need all that I can get. Can you explain the 3rd problem? I think the answer is 1/2...?