• Aug 14th 2009, 12:26 AM
bobbyboy1111
4^(x + 2) = 200

ln(4^(x+2) = ln(200)

xln(4) + 2ln(4) = ln(200)

xln(4) = -2ln(4) + ln(200)

xln(4)/ln(4) = (-2ln(4) + ln(200))*1/ln(4)

x = (-2ln(4) + ln(200))* 1/ln(4)

x = -2ln(4) + ln(200)/ln(4)

= 1.04

• Aug 14th 2009, 12:45 AM
QM deFuturo
Quote:

Originally Posted by bobbyboy1111

x = (-2ln(4) + ln(200))* 1/ln(4)

x = -2ln(4) + ln(200)/ln(4)

If you are going to divide the RHS by ln(4), then the -2ln(4) term becomes just -2.

You removed the parenthesis but didn't cancel out that term.

Btw, the easier way to solve this is just

log_4 (200) = x + 2

Then use change of base formula to compute log_4 (200). You'll get the same answer either way, but save a few steps.
• Aug 14th 2009, 01:15 AM
Twig
Hi!

$\displaystyle 4^{x+2}=200$

$\displaystyle (x+2)\cdot ln(4)=ln(200)$

$\displaystyle x\cdot ln(4)=ln(200)-2\cdot ln(4)$

$\displaystyle \Rightarrow \; x = \frac{ln(200)-2\cdot ln(4)}{ln(4)}$

Note: This can be simplified etc, but I didnt bother there, hoping it is easier to follow.