Show that -ln(4) = ln(0.25)???

**Henryt999** · January 10th 2010, 07:57 AM

How can I show that -ln(4) = ln(0.25).
A hint would be fine...

**VonNemo19** · January 10th 2010, 08:00 AM

Originally Posted by Henryt999

How can I show that -ln(4) = ln(0.25).
A hint would be fine...

Hint

Spoiler:

**earboth** · January 10th 2010, 08:01 AM

Originally Posted by Henryt999

How can I show that -ln(4) = ln(0.25).
A hint would be fine...

Add a zero to your equation:

$-\ln(4) = 0-\ln(4) = \ln(1)-\ln(4)$

I'll leave the rest for you.

**Dinkydoe** · January 10th 2010, 08:02 AM

It's a basic property of logarithms: $a\cdot \ln(b) = \ln(b^a)$

So if $a = -1, b= 4$ ? what is $b^{a}$

**Henryt999** · January 10th 2010, 08:07 AM

That was very very easy ops ops ops

**Soroban** · January 10th 2010, 08:49 AM

Hello, Henryt999!

How can I show that: . $-\ln(4) \:=\: \ln(0.25)$

The left side is: . $-1\cdot\ln(4) \;=\;\ln\left(4^{-1}\right) \;=\;\ln\left(\frac{1}{4}\right) \;=\;\ln(0.25)$

Thread: Show that -ln(4) = ln(0.25)???