How do you solve #3^x + 4 = 7^x - 1 #?

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Answer 1 · 2016-10-08T20:01:49

#x approx 1.08777#

Calling

#f(x) = 7^x-3^x-5# Expanding in Taylor series around #x_0#

#f(x) approx f(x_0)+((df)/(dx))_(x_0)(x-x_0)#

or

#f(x_1) approx f(x_0)+((df)/(dx))_(x_0)(x_1-x_0)#

If #x_0# is near a function zero, then #f(x_1) approx 0# so

#x_(k+1)=x_k-f(x_k)/((df)/(dx))_(x_k)#

Here

# ((df)/(dx))_(x_k)=7^(x_k) Log(7)-3^(x_k) Log(3) #

Begining with #x_0 = 1# we have

#x_0=1#
#x_1=1.09685#
#x_2=1.08786#
#x_3=1.08777#
#x_4=1.08777#