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

Printable View

- May 17th 2007, 05:31 PMtrancefanaticlogs
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 - May 17th 2007, 06:25 PMLogarithm Freak
*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 - May 17th 2007, 07:58 PMSoroban
Hello, trancefanatic!

Quote:

Evaluate: .log2(1/16)

Quote:

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

Quote:

Solve for x: .2·log3(x) - log3(2) .= .log3(8)

I got: x = 4 . Yes!

. . . . . . . .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