1. ## College 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

2. Originally Posted by Peyton Sawyer
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
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?

3. Originally Posted by Prove 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?
okay sorry.

there are like 40 more of these and these ones are the hardest for me and ive keept them aside for a while now, i was just hoping someone could help.

4. Hello Peyton Sawyer
Originally Posted by Peyton Sawyer
solve following equations...
Here are some hints. See if you can continue.
2ln3x=8
$\displaystyle \ln(3x)=4$

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

9.5e^.005x=19
$\displaystyle e^{0.005x}=2$

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

10^x^2=10^3x-2
$\displaystyle 10^{x^2}=10^{3x-2}$

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

6^x+1=4^2x-1
I assume this means:
$\displaystyle 6^{x+1}=4^{2x-1}$

$\displaystyle \Rightarrow (x+1)\log 6 = (2x-1)\log 4$ (where the log is to any base you like)
Collect like terms to solve for $\displaystyle x$ in the usual way.

lnx=3lnx=4
Doesn't make sense.

log3 (1-x)=1
Does this mean:
$\displaystyle \log_3 (1-x)=1$?
If so:
$\displaystyle 1-x = 3^1$

$\displaystyle \Rightarrow \text ?$
log(x+1) + log(x-1)=log3
$\displaystyle \log(x+1)(x-1)= \log 3$

$\displaystyle \Rightarrow \text ?$