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
Hello, trancefanatic!
We have: .log2(1/16) .= .log2(1/2^4) .= .log2(2^{-4}) .= .-4Evaluate: .log2(1/16)
Take logs of both sides: .ln[e^{x-2}] .= .ln(10)Solve for x: .e^{x-2} .= .10
i got: x = 4.30 . Right!
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
We have: .log3(x˛) - log3(2) .= .log3(8)Solve for x: .2·log3(x) - log3(2) .= .log3(8)
I got: x = 4 . Yes!
. . . . . . . .log3(x˛) .= .log3(8) + log3(2) .= .log3(8·2) .= .log3(16)
Then: .x˛ .= .16 . . → . . x .= .±4
But x = -4 is an extraneous root.
The only solution is: .x = 4