solve following equations...

2ln3x=8

9.5e^.005x=19

10^x^2=10^3x-2

6^x+1=4^2x-1

lnx=3lnx=4

log3 (1-x)=1

log(x+1) + log(x-1)=log3

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- Feb 26th 2010, 11:39 PMPeyton SawyerCollege Algebra
solve following equations...

2ln3x=8

9.5e^.005x=19

10^x^2=10^3x-2

6^x+1=4^2x-1

lnx=3lnx=4

log3 (1-x)=1

log(x+1) + log(x-1)=log3 - Feb 26th 2010, 11:48 PMProve It
1. Please do not post any more than 2 questions per thread. For the purposes of reading the answers given, the thread can become congested and difficult to keep track of. If you have more than 2 questions, create new threads for them (making sure not to post any more than 2 per thread). You are also more likely to get answers in a more timely manner.

2. Please show some effort. What have you tried? Exactly where are you getting stuck? - Feb 26th 2010, 11:53 PMPeyton Sawyer
- Feb 27th 2010, 12:46 AMGrandad
Hello Peyton SawyerHere are some hints. See if you can continue.

Quote:

2ln3x=8

$\displaystyle \Rightarrow 3x = e^{\text ?}$

Quote:

9.5e^.005x=19

$\displaystyle \Rightarrow 0.005x = \ln(\text ?)$

Quote:

10^x^2=10^3x-2

$\displaystyle \Rightarrow x^2=3x-2$

Quote:

6^x+1=4^2x-1

$\displaystyle 6^{x+1}=4^{2x-1}$Collect like terms to solve for $\displaystyle x$ in the usual way.

$\displaystyle \Rightarrow (x+1)\log 6 = (2x-1)\log 4$ (where the log is to any base you like)

Quote:

lnx=3lnx=4

Quote:

log3 (1-x)=1

$\displaystyle \log_3 (1-x)=1$?If so:$\displaystyle 1-x = 3^1$

$\displaystyle \Rightarrow \text ?$Quote:

log(x+1) + log(x-1)=log3

$\displaystyle \Rightarrow \text ?$

Grandad