Express in terms of natural logs...

**Savior_Self** · October 15th 2009, 10:06 AM

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

Thanks for your help.. I really appreciate it.

**HallsofIvy** · October 15th 2009, 10:13 AM

Originally Posted by Savior_Self

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

Thanks for your help.. I really appreciate it.

If $y= log_7(4)$ the $4= 7^y$ , from the definition of "log". Now solve for y by taking the natural logarithm of both sides.

**stapel** · October 15th 2009, 10:33 AM

Originally Posted by Savior_Self

Express $\log_{7}4$ in terms of natural logarithms.

Hint: Use the change-of-base formula.

**Savior_Self** · October 15th 2009, 10:54 AM

alright, I believe I've got it.

$x = \log_{7}4$

$7^x = 4$

$ln 7^x = ln 4$

$x ln 7 = ln 4$

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

all good?

**masters** · October 15th 2009, 10:57 AM

Originally Posted by Savior_Self

alright, I believe I've got it.

$x = \log_{7}4$

$7^x = 4$

$ln 7^x = ln 4$

$x ln 7 = ln 4$

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

all good?

All good!!

**Korupt** · October 15th 2009, 10:59 AM

Originally Posted by Savior_Self

alright, I believe I've got it.

$x = \log_{7}4$

$7^x = 4$

$ln 7^x = ln 4$

$x ln 7 = ln 4$

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

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:

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

It's pretty easy to use.

**e^(i*pi)** · October 15th 2009, 11:03 AM

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:

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

It's pretty easy to use.

Nay, your answer is a factor of -1 out.

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

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