Thread: Express in terms of natural logs...

1. Express in terms of natural logs...

Express $\displaystyle \log_{7}4$ in terms of natural logarithms. Do not find a numerical answer.

Thanks for your help.. I really appreciate it.

2. Originally Posted by Savior_Self
Express $\displaystyle \log_{7}4$ in terms of natural logarithms. Do not find a numerical answer.

Thanks for your help.. I really appreciate it.
If $\displaystyle y= log_7(4)$ the $\displaystyle 4= 7^y$, from the definition of "log". Now solve for y by taking the natural logarithm of both sides.

3. Originally Posted by Savior_Self
Express $\displaystyle \log_{7}4$ in terms of natural logarithms.
Hint: Use the change-of-base formula.

4. alright, I believe I've got it.

$\displaystyle x = \log_{7}4$

$\displaystyle 7^x = 4$

$\displaystyle ln 7^x = ln 4$

$\displaystyle x ln 7 = ln 4$

$\displaystyle x = \frac{ln4}{ln7}$

all good?

5. Originally Posted by Savior_Self
alright, I believe I've got it.

$\displaystyle x = \log_{7}4$

$\displaystyle 7^x = 4$

$\displaystyle ln 7^x = ln 4$

$\displaystyle x ln 7 = ln 4$

$\displaystyle x = \frac{ln4}{ln7}$

all good?
All good!!

6. Originally Posted by Savior_Self
alright, I believe I've got it.

$\displaystyle x = \log_{7}4$

$\displaystyle 7^x = 4$

$\displaystyle ln 7^x = ln 4$

$\displaystyle x ln 7 = ln 4$

$\displaystyle x = \frac{ln4}{ln7}$

all good?
Yes, but there is no point in doing all that. With the change of base formula (as stapel pointed out) you can change between log easy. Change of base says:

$\displaystyle \log_x y = \frac{\log_z x}{\log_z y}$

It's pretty easy to use.

7. Originally Posted by Korupt
Yes, but there is no point in doing all that. With the change of base formula (as stapel pointed out) you can change between log easy. Change of base says:

$\displaystyle \log_x y = \frac{\log_z x}{\log_z y}$

It's pretty easy to use.

$\displaystyle log_x y = -\frac{log_z x}{log_z y} = \frac{log_z y}{log_z x}$