Need help w/ ?log?_2 (4^(x-1)-1)=?log?_2 15+x-3

[oded244]

in the first one i get 10 and 1/10 but my book says i need to get 10 and1/100,
i need help with the second one.

Thanks

 

[skeeter]

log(10x)*log(x) = 2

[log(10) + log(x)]*log(x) = 2

[1 + log(x)]*log(x) = 2

[log(x)][sup:3uem2fm2]2[/sup:3uem2fm2] + log(x) - 2 = 0

[log(x) + 2][log(x) - 1] = 0

log(x) = -2 or log(x) = 1

see where the 1/100 comes from?


log[sub:3uem2fm2]2[/sub:3uem2fm2](4[sup:3uem2fm2]x-1[/sup:3uem2fm2] - 1) = log[sub:3uem2fm2]2[/sub:3uem2fm2]15 + x - 3

4[sup:3uem2fm2]x-1[/sup:3uem2fm2] - 1 = 2[sup:3uem2fm2]log[sub:3uem2fm2]2[/sub:3uem2fm2]15 + x - 3[/sup:3uem2fm2]

4[sup:3uem2fm2]x-1[/sup:3uem2fm2] - 1 = 2[sup:3uem2fm2]log[sub:3uem2fm2]2[/sub:3uem2fm2]15[/sup:3uem2fm2] * 2[sup:3uem2fm2]x - 3[/sup:3uem2fm2]

4[sup:3uem2fm2]x-1[/sup:3uem2fm2] - 1 = 15 * 2[sup:3uem2fm2]x - 3[/sup:3uem2fm2]

4[sup:3uem2fm2]x-1[/sup:3uem2fm2] - 1 = 15 * 2[sup:3uem2fm2]x - 1[/sup:3uem2fm2] * 2[sup:3uem2fm2]-2[/sup:3uem2fm2]

4[sup:3uem2fm2]x-1[/sup:3uem2fm2] - 1 = (15/4) * 2[sup:3uem2fm2]x - 1[/sup:3uem2fm2]

(2[sup:3uem2fm2]2[/sup:3uem2fm2])[sup:3uem2fm2]x-1[/sup:3uem2fm2] - 1 = (15/4) * 2[sup:3uem2fm2]x - 1[/sup:3uem2fm2]

(2[sup:3uem2fm2]x-1[/sup:3uem2fm2])[sup:3uem2fm2]2[/sup:3uem2fm2] - (15/4) * (2[sup:3uem2fm2]x-1[/sup:3uem2fm2]) - 1 = 0

let u = 2[sup:3uem2fm2]x-1[/sup:3uem2fm2] ...

u[sup:3uem2fm2]2[/sup:3uem2fm2] - (15/4)u - 1 = 0

4u[sup:3uem2fm2]2[/sup:3uem2fm2] - 15u - 4 = 0

(u-4)(4u+1) = 0

u = 4 ... u = -1/4

x = 3 ... no real solution

 

[stapel]

in the first one i get 10 and 1/10 but my book says i need to get 10 and1/100,
i need help with the second one.

In future (in case one of the two guys who "do" homework doesn't surf by that day... or that week....), please show what you've done. :!:

I'm afraid that, having only from a mention of your final results, the tutors are unable to "see" what you've done or where you might be going wrong. By showing your steps and reasoning, they'll be better able to help you correct any systemic errors, so you can succeed on your own! :wink:

Thank you!

Eliz.

 

[oded244]

Thanks , you helped me a lot
no problem Eliz, i'll do that.. sometimes i don't know how to start, though.
I have some more questions regarding logs, should i ask them here or do i need to open a new thread?

 

[stapel]

no problem...i'll do that.. sometimes i don't know how to start, though.

If you have no idea how even to get started, then say that, so the tutors can try to find lesson links. Once you've studied the topic, you'll be able to make an attempt, and we can help from there. :wink:

I have some more questions regarding logs, should i ask them here or do i need to open a new thread?

New questions are generally better put into new threads. Thank you!

Eliz.