How do you solve #7^(x – 2) = 12x#?

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Answer 1 · 2015-11-01T14:00:59

A step in the correct direction!! Perhaps someone else can take it further. I used the Iteration method.

#color(red)("Method 1")#
Plot the straight line graph of #12x#
Plot the curve of #7^(x-2)#
The intersections are where the values for #x# are.

Or try iteration for #x =7^x/588#

#color(red)("Method 2")#
#7^(x-2) -> (7^x)/(7^2)#

so #7^(x-2) = 12x -> 7^x = (7^2)(12)x#

#7^x = 588x#

Taking logs
#x ln(7) = ln(588) + ln(x)#

#x = (ln(588) + ln(x))/(ln(7))#

By iteration

Set seed value as 1
#x ~= 3.988# to 3 decimal places

I could not find the seed for the lower one which is in the region of
#x ~= 0.0016# to 5 dp

Graphs of the log method:
Tony B

Tony B