How do you solve #4(3^2x) = e^x#?

1 Answer
Tony B
Nov 9, 2015

Took it to a point and found by iteration 1 of the two solutions as
#~~5.23980.....# Perhaps some one else could take the method further to help you.

Explanation:

Given: #4(3^2x) = e^x#

~~~~~~~~~~~~~~ Pre amble ~~~~~~~~~~~~~~~
The key here is than #log_e (e)=1#

#log_e# is normally written as ln. So #log_e(e) =ln(e)=1#

The same way that #log_10(10)=1#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write as #" "36x =e^x #

Taking logs

#ln(36) + ln(x) =xln(e)#

#ln(36) +ln(x) =x#

By the way; #36 = 6^2 -> ln(32) =ln(6^2) = 2 ln(6)#

Using iteration with a seed value of 3 gives 1 potential solution of

#5.23980.....#

Plotting this as #y= ln(36)+ln(x)-x# shows that there are two solutions when y=0. I could not find the seed value for iteration solution at #x~~ #0.0285...#

http://www.wolframalpha.com/input/?i=2ln%286%29%2Bln%28x%29-x%3D0

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