College Algebra

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February 26th 2010, 11:39 PM
Peyton Sawyer

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
February 26th 2010, 11:48 PM
Prove It

Quote:

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?
February 26th 2010, 11:53 PM
Peyton Sawyer

Quote:

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.
February 27th 2010, 12:46 AM
Grandad

Hello Peyton Sawyer

Quote:

Originally Posted by Peyton Sawyer

solve following equations...

Here are some hints. See if you can continue.

Quote:

2ln3x=8

$\ln(3x)=4$

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

Quote:

9.5e^.005x=19

$e^{0.005x}=2$

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

Quote:

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

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

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

Quote:

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

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

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

Quote:

lnx=3lnx=4

Doesn't make sense.

Quote:

log3 (1-x)=1

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

$\Rightarrow \text ?$

Quote:

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

$\log(x+1)(x-1)= \log 3$

$\Rightarrow \text ?$

Grandad