exponential functions

**nando29** · July 29th 2010, 09:01 PM

I'm having trouble with the following:

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

**pickslides** · July 29th 2010, 09:08 PM

Can i suggest

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

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

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

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

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

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

can you finish it?

**nando29** · July 30th 2010, 04:49 PM

yeah it's very helpful!

my ans:

ln4 / 7ln4 + ln5

**pickslides** · July 30th 2010, 06:29 PM

be careful with how you write your answer

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

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

Thread: exponential functions

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