Hey guys I am new to this website and wanted some help. I would like to see the steps if possible. Thank you!

Solve for t or x. Round off 3 significant decimals

ln(t/t-2)=1

5^(t+1)=6^t

ln9x=1

2=e^(.6t)

1/5=e^(4x)

9ln x = 15

ln 3 + ln x= 1.4

Printable View

- Apr 12th 2010, 11:53 PMJackie10ln or log problems im confused
Hey guys I am new to this website and wanted some help. I would like to see the steps if possible. Thank you!

Solve for t or x. Round off 3 significant decimals

ln(t/t-2)=1

5^(t+1)=6^t

ln9x=1

2=e^(.6t)

1/5=e^(4x)

9ln x = 15

ln 3 + ln x= 1.4 - Apr 13th 2010, 12:03 AMDebsta
- Apr 13th 2010, 12:05 AMProve It
$\displaystyle \ln{\frac{t}{t - 2}} = 1$

$\displaystyle \frac{t}{t - 2} = e^1$

$\displaystyle \frac{t - 2 + 2}{t - 2} = e$

$\displaystyle 1 + \frac{2}{t - 2} = e$

$\displaystyle \frac{2}{t - 2} = e - 1$

$\displaystyle \frac{t - 2}{2} = \frac{1}{e - 1}$

$\displaystyle t - 2 = \frac{2}{e - 1}$

$\displaystyle t = \frac{2}{e - 1} + 2$

$\displaystyle t = \frac{2}{e - 1} + \frac{2(e - 1)}{e - 1}$

$\displaystyle t = \frac{2 + 2e - 2}{e - 1}$

$\displaystyle t = \frac{2e}{e - 1}$. - Apr 13th 2010, 09:48 AMearboth
1. I assume that you are familiar with the basic laws of powers and logarithms.

2.

$\displaystyle 5^{t+1}=6^t$

$\displaystyle 5 \cdot 5^t=6^t$

$\displaystyle 5=\frac{6^t}{5^t}=\left(\frac65 \right)^t$

$\displaystyle t=\log_{\frac65}(5)=\frac{\ln(5)}{\ln(6)-\ln(5)}$ - Apr 13th 2010, 09:51 AMe^(i*pi)
- Apr 13th 2010, 03:40 PMJackie10Thank you!
So for 2=e^.6t

do i take

ln of both sides

ln2=lne^.6t

then

lne=1

ln2=.6931

.6931=.6t

to get t i divide by .6 to both sides

then i get

.6931/.6

t=1.155 - Apr 13th 2010, 05:22 PMharish21