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September 1, 2014

September 1, 2014

Posted by **anonymous** on Saturday, February 5, 2011 at 3:21am.

2^(9x-7)=7

My first step was to change it to

log(base 2) 7 = 9x-7

i got 1.0897

Did i do something wrong or did i miss one question?

- pre calculus -
**Reiny**, Saturday, February 5, 2011 at 7:02amjust take log of both sides

log (2^(9x-7)) = log 7

(9x-7) log2 = log7

9x - 7 = log7/log2

etc.

your answer is correct, see

http://www.google.ca/search?hl=en&source=hp&q=%28%28log%287%29%2Flog%282%29%2B7%29%2F9&btnG=Google+Search&aq=f&aqi=&aql=&oq=

- pre calculus -
**drwls**, Saturday, February 5, 2011 at 7:03am[9x -7] log 2 = log 7

Any log base can be used, as long as it is the same for both sides.

9x - 7 = 2.807

9x = 9.807

x = 1.0897

Your answer is correct!

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