X^(log_10(x) + 2) = 1000

Please include detail on how you got the solution (which I know is 10). Thanks.

x^( log x + 2 ) = 1000
(log x + 2 )log x = log 1000
(log x )^2 + 2 log x = 3
( log x )^2 + 2 log x - 3 = 0
( log x + 3 )(log x - 1 ) = 0
log x = - 3 or log x = 1
x = 1 / 1000 or 10

x^(log(x) +2) = x^(log(x)) * x²

1000 = 10³

so we have:

x^(log(x)) * x² = 10³

There is no analytical method for solving this equation, it must be done by inspection.

By definition: log_10(x) = y where 10^y = x

In this equation we have x^y * x² = 10³ so it would be really helpful if y =1

giving us x³ = 10³, from which x is clearly 10

well if x is 10, y actually does = 1:

log_10(10) = 1

It really is just by inspection, once you get used to using logs you'll start thinking of everything in terms of powers and the connections become more obvious

x^(log_{10}(x) + 2) = 1000
log_{10}(x^(log_{10}(x) + 2)) = log_{10}(10^3)
(log_{10}(x) + 2)•log_{10}(x) = 3
(ln(x)/ln(10) + 2)•ln(x)/ln(10) = 3
(ln(x) + 2•ln(10))•ln(x)/(ln(10)^2) = 3
ln(x)^2 + 2•ln(10)•ln(x) = 3•ln(10)^2
ln(x)^2 + 2•ln(10)•ln(x) - 3•ln(10)^2 = 0
ln(x)^2 - 1•ln(10)•ln(x) + 3•ln(10)•ln(x) - 3•ln(10)^2 = 0
ln(x)•(ln(x) - ln(10)) + 3•ln(10)•(ln(x) - ln(10)) = 0
(ln(x) - ln(10))(ln(x) + 3•ln(10)) = 0
=> ln(x) - ln(10) = 0, ln(x) = ln(10), e^ln(x) = e^ln(10), x = 10
=> ln(x) + 3•ln(10) = 0, ln(x) = -3•ln(10), ln(x) = ln(10^(-3)), e^ln(x) = e^ln(10^(-3)), x = 10^(-3)
So x ∈ {.001,10}.