What are the solutions of 100^(7x+1) = 1000^(3x-2)1007x+1=10003x−2 ?

1 Answer
George C.
Mar 12, 2017

Real valued solution:

x=-8/5x=−85

Complex valued solutions:

x = -8/5+(2kpii)/(5ln(10))" "x=−85+2kπi5ln(10) for any integer kk

Explanation:

Note that if a > 0a>0 then:

(a^b)^c = a^(bc)(ab)c=abc

So we find:

10^(14x+2) = 10^(2(7x+1)) = (10^2)^(7x+1) = 100^(7x+1)1014x+2=102(7x+1)=(102)7x+1=1007x+1

10^(9x-6) = 10^(3(3x-2)) = (10^3)^(3x-2) = 1000^(3x-2)109x−6=103(3x−2)=(103)3x−2=10003x−2

So if:

100^(7x+1) = 1000^(3x-2)1007x+1=10003x−2

then:

10^(14x+2) = 10^(9x-6)1014x+2=109x−6

Then either say "take common logarithms of both sides" or simply note that as a real-valued function 10^x10x is one to one to deduce:

14x+2 = 9x-614x+2=9x−6

Subtract 9x+29x+2 from both sides to get:

5x=-85x=−8

Divide both sides by 55 to get:

x = -8/5x=−85

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Footnote

There are also Complex valued solutions, which we can deduce from Euler's identity:

e^(ipi)+1 = 0eiπ+1=0

Hence:

e^(2kpii) = 1" "e2kπi=1 for any integer kk

Then:

10 = e^(ln 10)10=eln10

So:

10^x = e^(x ln 10)10x=exln10

So we find:

10^((2kpii)/ln(10)) = 1" "102kπiln(10)=1 for any integer kk

Hence we find:

10^(14x+2) = 10^(9x-6)1014x+2=109x−6

if and only if:

14x+2 = 9x-6 + (2kpii)/ln(10)" "14x+2=9x−6+2kπiln(10) for some integer kk

Hence:

5x = -8+(2kpii)/ln(10)5x=−8+2kπiln(10)

So:

x = -8/5+(2kpii)/(5ln(10))x=−85+2kπi5ln(10)

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