hi, me again, how to solve this, logx + lnx = 10, does anyone know? thank you for your time and help. EDIT: also how do logarithms help to solve simultaneous eqautions, such as x+y=5 and xy=6, or xy=24 and (e^x)(e^y)=e^10 etc. thanks anyone
hi, me again, how to solve this, logx + lnx = 10, does anyone know? thank you for your time and help. EDIT: also how do logarithms help to solve simultaneous eqautions, such as x+y=5 and xy=6, or xy=24 and (e^x)(e^y)=e^10 etc. thanks anyone
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 .[1]Solve: .xy .= .24, .(e^x)(e^y) .= .e^10
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)