Hello, captainlewis!

Use the Base-Change formula . . .Solve: .log(x) + ln(x) .= .10

. . .ln(x)

. . ------- + ln(x) .= .10

. . ln(10)

Multiply by ln(10): .ln(x) + ln(10)·ln(x) .= .10·ln(10)

Factor: . ln(x)·[1 + ln(10)] .= .10·ln(10)

. . . . . . . . . . . . .10·ln(10)

Then: . ln(x) .= .------------ .≈ .6.97

. . . . . . . . . . . . 1 + ln(10)

Therefore: .x .= .e^6.97 .≈ .1064.2

The second equation is: .e^{x+y} .= .e^10 . → . x + y .= .10 .Solve: .xy .= .24, .(e^x)(e^y) .= .e^10[1]

The first equation is: .xy = 24 . → . y = 24/x .[2]

Substitute [2] into [1]: .x + 24/x .= .10

. . which gives us a quadratic: .x² - 10x + 24 .= .0

. . which factors: .(x - 4)(x - 6) .= .0

. . and has roots: .x .= .4, 6

Substitute into [2]: .y .= .6, 4

Therefore: .(x,y) .= .(4, 6) and (6, 4)