Logarithms

**JennaK** · October 12th, 2009, 06:10 PM

I've gotten this far so far I just don't know what to do and how to finish. Any help is appreciated

log2 (x+1) = 1 +log4 (3x-2)
2[log2(x+1) = 1+ log4 (3x-2)]
2 log2 (x+1) = 2 +log2 ( 3x-2)
log2 (x+1)^2 - log2 ( 3x-2 ) =2
log2 (x+1)^2 / 3x-2 =2
log2 ( 2x+2)/3x-2 =2
and this is where I get stuck. Any ideas?

**pickslides** · October 12th, 2009, 07:04 PM

Originally Posted by JennaK

log2 (x+1) = 1 +log4 (3x-2)
2[log2(x+1) = 1+ log4 (3x-2)]
2 log2 (x+1) = 2 +log2 ( 3x-2)

Can you explain this? It makes no sense to me.

I would try

$\log_2 (x+1) = 1 +\log_4 (3x-2)$

$\log_2 (x+1) = \log_44 +\log_4 (3x-2)$

$\log_2 (x+1) = \log_44 +\log_4 (3x-2)$

$\log_2 (x+1) = \log_4 4(3x-2)$

$\log_2 (x+1) = \log_4 (12x-8)$

Do you follow?

**JennaK** · October 12th, 2009, 07:17 PM

My teacher came up with the other part of it and I wasnt sure how she got that part either. Yes, I think I follow it. I'm just not too sure how to change bases, would I multiply the base I want to change by each side?

Thankyou

Math Help Forum: Logarithms

LinkBack

Thread Tools

Display

Logarithms

Similar Math Help Forum Discussions

logarithms help

Logarithms

logarithms

Logarithms

logarithms