# ln or log problems im confused

• Apr 12th 2010, 11:53 PM
Jackie10
ln 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 AM
Debsta
Quote:

Originally Posted by Jackie10
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

I'll take you through them one at a time if you like, but first show me how you would start Q1.
• Apr 13th 2010, 12:05 AM
Prove It
Quote:

Originally Posted by Jackie10
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

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

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

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

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

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

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

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

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

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

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

$\displaystyle t = \frac{2e}{e - 1}$.
• Apr 13th 2010, 09:48 AM
earboth
Quote:

Originally Posted by Jackie10
Hey guys I am new to this website and wanted some help. I would like to see the steps if possible. <<<<< we also
Thank you!

Solve for t or x. Round off 3 significant decimals
...
5^(t+1)=6^t
...

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 AM
e^(i*pi)
Quote:

Originally Posted by Jackie10
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

$\displaystyle \ln(9x) = 1$

$\displaystyle 9x = e^1 = e\$

$\displaystyle x = \frac{e}{9}$
• Apr 13th 2010, 03:40 PM
Jackie10
Thank 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 PM
harish21
Quote:

Originally Posted by Jackie10
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

Correct!!(Clapping)