Question

how do u solve log of x 5=1/4

Answer 1

You need to be a little clearer. Is it,\[\log_x5=\frac{1}{4}\]or\[\log x^5=1/4\]?

Answer 2

1st

Answer 3

ok

Answer 4

one sec...dealing with a couple of things at once.

Answer 5

You have to use the definition of what the logarithm means. If you have\[y=a^x\]then by definition of the logarithm,\[\log _a y = x\]
By the definition then, you have\[\log_x 5 = 1/4 \rightarrow x^{1/4}=5\]Raise both sides to the power of 4, and you have\[(x^{1/4})^4=5^4 \rightarrow x=625\]

Answer 6

i have another u could help me with

Answer 7

and thanks

Answer 8

ok baseballkid, but I could use another fan :p

Answer 9

ln(5x-3)+ln2- ln(24-2x)

Answer 10

how do i become your fan

Answer 11

You have to know your log laws. The two you need here are:\[\log a + \log b = \log ab\]and\[\log a - \log b = \log \frac{a}{b}\]Here, for the frist two terms, you have\[\ln ( 5x-3)+\ln 2 = \ln (2(5x-3))=\ln (10x-6)\]and then\[\ln (10x-6) -\ln (24-2x)= \ln \frac{10x-6}{24-2x}=\ln \frac{5x-3}{12-x}\]since the numerator and denominator have a common factor of 2.

Answer 12

ok im an idiot i put it in wrong i meant to put ln(5x-3) + ln2 = ln(24-2x)

Answer 13

but why r u so smart? i mean i think its great

Answer 14

Okay. Now you're solving for x...that's different.

Answer 15

lol, so smart...thanks...

Answer 16

You need to exponentiate both sides of your equation:\[e^{\ln(5x-3)+\ln2}=e^{\ln(24-2x)} \rightarrow e^{\ln2(5x-3)}=e^{\ln(24-2x)}\]\[\rightarrow 2(5x-3)=24-2x\]

Answer 17

You expand and solve for x now.

Answer 18

Do you see what I did with exponentiation? It 'undoes' the logarithm. It's the 'inverse' of logarithm.

Answer 19

Once you are done with this, could you please provide me some help

Answer 20

I may have to just take a look and go. I'm supposed to be elsewhere today.

Answer 21

Ok it will only take a look from a genius like you
Here it is
http://openstudy.com/groups/mathematics#/updates/4da834d1d6938b0bf8d5a44d

Answer 22

baseballkid, are you okay with this? If not, leave a message here and when I can, I'll get back to you.

Answer 23

ok thank you so much. i have some more ill post later

Answer 24

ok

Answer 25

post them for others in the question box if you need them quickly (others will do them), else, post here and I'll do what I can. :)

Answer 26

\[\log_{5} (3x+10)-3\log_{5}4=2 \]

Answer 27

You have to use the log laws,\[\log_5 (3x+10)-3\log _ 5 4 =\log_5 (3x+10)-\log _ 5 4 ^3\]\[=\log 5 \frac{3x+10}{64}\]that is,\[\log _ 5 \frac{3x+10}{64}=2 \rightarrow \frac{3x+10}{64}=5^2\]by definition of the logarithm. Solving for x, you have\[3x+10=25 \times 64 \rightarrow x = 530\]