Question

\[\log_{7} 343+\log_{2} \sqrt{32}-\log_{1/2} (1/2)\]

Answer 1

\[\LARGE \log_{n}{x} = \frac{\log{x}}{\log{n}}\]
\[\Large \log_{7}{343} +\log_{2}{\sqrt{32}} -\log_{(1/2)}{1/2} = \frac{\log{343}}{\log{7}}+\frac{\log{\sqrt{32}}}{\log{2}}-\frac{\log{1/2}}{\log{1/2}}\]

Answer 2

write all those numbers, inside the logarithms, as the base of the logarithm, raised to some power. Then it's pretty easy to work out what it comes out as.

Answer 3

the answer is 9/2

Answer 4

@nitz please don't just give out the answer. That's against the code of conduct.

Answer 5

\[\log(7^{3})/\log7+\log \sqrt{2^{5}}/\log 2-\log(1/2)/\log(1/2)\]

Answer 6

I think they want you to use factorization on those inside numbers. So for example 343 = 7^3. Then, log_{a} {b} essentially is asking to solve a^x = b for x. So for that first term, you need to solve 7^x = 343. And I just said 7^3 = 343.

Also, for the second term you can simplify that a bit and put it as 2^x = sqrt(32)

Answer 7

7^3 = 343 and \(\log_nM^a = a\log_nM\)

use these two fundas to solve^.

Answer 8

now you can use \[\log(m ^{n})=nlog(m)\]

Answer 9

\[\log_{7} 343=x
7^{x} = 7^{3}
x=3\]

\[\log_{2}\sqrt{343} =x
2^{x} =2^{5}
x=5\]

\[\log_{1/2} (1/2)=x
1/2^{x} = 1/2
x=1/2\]

3+5-1/2=7/2
is it right??
@nitz
@apoorvk
@scarydoor

Answer 10

first line yes, next two lines, no.

Answer 11

actually I don't know what you're doing with those x's, but the first and last part of the first line are right.

Answer 12

sorry is \[\sqrt{32}\] not 343

Answer 13

\[\log(\sqrt{32})/\log(2)=\log(\sqrt{2^{5}})/(\log(2))=(5/2)*(\log 2)/(\log2)=5/2\]

Answer 14

32 = 2*16 = 2*2^2. So sqrt(32) = 2*sqrt(2).

log_2 (sqrt(32)) is equivalent to asking 2^x = sqrt(32). and sqrt(32) = 2*sqrt(2) = 2^(3/2).

log_(1/2) (1/2) = x is equivalent to asking (1/2) ^ x = (1/2). x is clearly equal to 1.

Answer 15

\[\log_{1/2} (1/2)\]

how to solve @nitz

Answer 16

sorry my first line is wrong. 16 = 4^2 = 2^4 ... so adjust the numbers for that.

Answer 17

\[\log _{n}(m)=\log(m)/\log(n)\]
\[\log(1/2)/\log(1/2)=1\]

Answer 18

I suggest just remembering what the logarithm function was invented for. You had exponential functions where you have a^x = y, for some base a. Then people wanted to take the inverse of that. What I mean is that if they knew that a^x = y and they knew both a and y, they wanted a function that worked the other way. This function is log_a (y) = x. You just need to figure out "a to the power [something] equals y". since you can factor all the y's in terms of a, then it's easy to work out and you don't have to bother with any kind of change of base. I think that actually makes it harder in this case.

Answer 19

and think about "half to the power [something] is equal to a half" what is the "something"? it clearly has to be one. Messing around with it any more is just making it more complicated for yourself I think.

Answer 20

sometimes the notation of the logarithm function confuses people away from the simplicity of what it's asking...

Answer 21

ln and log is the same ?

Answer 22

I think there are different standards. Where I am, ln is used when the base is the number e. ln is then called the natural logarithm, because e is used a lot in "natural" things / science. ln x = log_e x.

Answer 23

ln-natural log with base e
log -with base 10

Answer 24

occasionally log is used without mentioning the base when the base is 2. It depends on the people who are using the maths usually... computer scientists often use log to mean base 2 unless otherwise specified. Probably base 10 is more common though, if the base isn't specified.

Answer 25

k thanks :) All
@nitz
@scarydoor
@Wired
@apoorvk

Answer 26

for example this quest
ln x= ln17+ ln13
how to solve with ln

Answer 27

Use the funda --> \(log_a m + \log_a n = =\log_a(mn)\)

Answer 28

*single '=' sign.

And yes, then you remove the log signs (since bases are same).

Answer 29

e is irrational. So, I'm pretty sure (haven't done this stuff in a while) that you can only solve the right hand side by using a calculator, or, at if you for some reason are told not to, then by a numerical method that will approximate it to some desired accuracy, which is what the calculator would essentially do. But the teacher wouldn't be asking that...

So you solve right hand side. Say ln x = y, where you know what y is. Then raise e to the power of both sides, so you have e^(ln x) = x = e ^ y. Then you again have to use the calculator to solve e ^ y, and then you've solved for x.

Answer 30

ln(17+13)=30
is it right?

Answer 31

apoorvk's response above is better than mine. I forgot you can combine them and then not bother about calculating any logarithms.

Answer 32

\[\log_{7} 343+\log_{2} \sqrt{32}-\log_{1/2} (1/2)\]\[\implies \log_7{7^3} +\log_2{2^{5*\frac{1}{2}}} -\log_{2^{-1}}{2^{-1}}.\]\[\implies 3\log_7{7}+5*\frac{1}{2}\log_2{2}-(\frac{-1}{-1})\log_2{2} .\]\[\implies 3+\frac{5}{2}-1=?\]It is becoz \[\log_a{a}=1.\]

Answer 33

do this urself now:)

Answer 34

by what apoorvk said above, ln 17 + ln 13 = ln(17*13).

Then left hand side and right hand side are the same number. So take the number e and raise that by the number on left hand side, and that's equal to e raised to the number on right hand side.

Then remember that e ^ ln(x) = x. To see why that's true, remember what the ln function does. it's an inverse of e^x function.

So then you have x = 17 * 13. I think. Unless I said something silly.

Answer 35

@nitz Please do refrain from giving out answers. Thanks.