I'm having trouble with the following:

Attachment 18382

i cannot find a problem similar in my book and my friend is stump too. there is an option for no solution. (Bow)

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- Jul 29th 2010, 09:01 PMnando29exponential functions
I'm having trouble with the following:

Attachment 18382

i cannot find a problem similar in my book and my friend is stump too. there is an option for no solution. (Bow) - Jul 29th 2010, 09:08 PMpickslides
Can i suggest

$\displaystyle 4^{1-7x}= 5^{x} $

$\displaystyle \ln 4^{1-7x}= \ln 5^{x} $

$\displaystyle (1-7x)\ln 4= x\ln 5 $

$\displaystyle \ln 4-7x\ln 4= x\ln 5 $

$\displaystyle \ln 4= x\ln 5+7x\ln 4 $

$\displaystyle \ln 4= x(\ln 5+7\ln 4) $

can you finish it? - Jul 30th 2010, 04:49 PMnando29
yeah it's very helpful!

my ans:

ln4 / 7ln4 + ln5 - Jul 30th 2010, 06:29 PMpickslides
be careful with how you write your answer

**ln4 / 7ln4 + ln5**could imply only $\displaystyle \frac{\ln 4 }{7\ln 4}+\ln5 = \frac{1 }{7}+\ln5$

To stop any confusion use brackets like this**ln4 / (7ln4 + ln5)**