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

##### 1 Answer

Real valued solution:

#x=-8/5#

Complex valued solutions:

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

#### Explanation:

Note that if

#(a^b)^c = a^(bc)#

So we find:

#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

#14x+2 = 9x-6#

Subtract

#5x=-8#

Divide both sides by

#x = -8/5#

**Footnote**

There are also Complex valued solutions, which we can deduce from 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)#

So we find:

#10^((2kpii)/ln(10)) = 1" "# for any integer#k#

Hence we find:

#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))#