What are the solutions of #100^(7x+1) = 1000^(3x-2)# ?

Question

James Allen · Accepted Answer

Real valued solution:

#x=-8/5#

Complex valued solutions:

#x = -8/5+(2kpii)/(5ln(10))" "# for any integer #k#

Note that if #a > 0# then: #(a^b)^c = a^(bc)# Thus, we discover: #10^(14x+2) = 10^(2(7x+1)) = (10^2)^(7x+1) = 100^(7x+1)# #10^(9x-6) = 10^(3(3x-2)) = (10^3)^(3x-2) = 1000^(3x-2)# So if: #100^(7x+1) = 1000^(3x-2)# then: #10^(14x+2) = 10^(9x-6)# Then either say "take common logarithms of both sides" or simply note that as a real-valued function #10^x# is one to one to deduce: #14x+2 = 9x-6# Subtract #9x+2# from both sides to get: #5x=-8# Divide both sides by #5# to get: #x = -8/5# #color(white)()# Footnote Additionally, Complex valued solutions exist, as shown by Euler's identity: #e^(ipi)+1 = 0# Hence: #e^(2kpii) = 1" "# for any integer #k# Then: #10 = e^(ln 10)# So: #10^x = e^(x ln 10)# Thus, we discover: #10^((2kpii)/ln(10)) = 1" "# for any integer #k# Thus, we discover: #10^(14x+2) = 10^(9x-6)# if and only if: #14x+2 = 9x-6 + (2kpii)/ln(10)" "# for some integer #k# Hence: #5x = -8+(2kpii)/ln(10)# So: #x = -8/5+(2kpii)/(5ln(10))#

Sadie Ackerman · Answer

undefined The solutions of the equation 100^(7x + 1) = 1000^(3x - 2) are x = -0.2 and x = 0.2.