How do you solve the exponential equation #100^(7x+1)=1000^(3x-2)#?

2 Answers
Mar 13, 2017

#x = -8/5#

Explanation:

In an exponential equation: #"base"^"exponent"#

Make both #100# and #1000# be a power of #10# so they both have the same #"base"#

#100 = 10^2# and #1000 = 10^3#

Use the exponent rule power of a power #(x^m)^n = x^(m*n)#

#(10^2)^(7x+1) = (10^3)^(3x-2)#

Simplify the exponents:

#10^(14x+2) = 10^(9x-6)#

When the bases are equivalent, the exponents are equivalent:
#14x+2 = 9x-6#

Solve for #x# by adding/subtracting like-terms:
#5x = -8#

So #x = -8/5#

Mar 13, 2017

I tried this:

Explanation:

Let us take the log in base #10# of both sides and use properties of logs:

#log_(10)100^(7x+1)=log _(10)1000^(3x-2)#

#(7x+1)log_(10)100=(3x-2)log_(10)1000#

#(7x+1)*2=(3x-2)*3#

#14x+2=9x-6#

#5x=-8#

#x=-8/5#