Thread: logs

are my answers right? they're due tomorrow

evaluate: log(2)1/16
i got -2.3344

solve for x: e^(x-2)=10
i got 4.30 = x

solve for x: 2log(3)x - log(3)2 = log(3)8
i got x = 4

evaluate: log(2)1/16
i got -2.3344

solve for x: e^(x-2)=10
i got 4.30 = x

solve for x: 2log(3)x - log(3)2 = log(3)8
i got x = 4

(note: I'm assuming that the (2) is for base 2)

(Log1/16)/(Log2)
-4


Ln(e^x-2)= ln10
x-2= ln10
x-2+2= 2.3025 + 2
x= 4.3025

Hello, trancefanatic!

Evaluate: .log2(1/16)

We have: .log2(1/16) .= .log2(1/2^4) .= .log2(2^{-4}) .= .-4

Solve for x: .e^{x-2} .= .10
i got: x = 4.30 . Right!

Take logs of both sides: .ln[e^{x-2}] .= .ln(10)

Then we have: .(x-2)·ln(e) .= .ln(10)

Since ln(e) = 1, we get: .x - 2 .= .ln(10)

Therefore: .x .= .ln(10) + 2 .≈ .4.30

Solve for x: .2·log3(x) - log3(2) .= .log3(8)
I got: x = 4 . Yes!

We have: .log3(x²) - log3(2) .= .log3(8)

. . . . . . . .log3(x²) .= .log3(8) + log3(2) .= .log3(8·2) .= .log3(16)

Then: .x² .= .16 . . → . . x .= .±4

But x = -4 is an extraneous root.
The only solution is: .x = 4