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

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

Don't forget the exponent is subtracted when the former is split via the latter. Don't try to mislead others. In this case common factor is to be taken and then simplified. 2^one thousand - 2^999 = 2^999 (2 - 1) = 2^999 * 1 = 2^999

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}.