Log problems....

[Kanjiru]

I understand the basic log functions, but I don't understand how to do these problems, help would be greately appreciated...thanx.

1)2^(logx)=4

2)5e^(x)=e^(3x=+1)

 

[pka]

The equation 2ⁿ=4 has a solution n=2.
So that 2^log(x)=4 gives log(x)=2.
What is the solution ?

 

[soroban]

Hello, Kanjiru!

1) 2^{log x} = 4
.

Since 4 = 2², we have: .2^{log x} .= .2²

Equating exponents: .log x .= .2 . ---> .x .= .10² .= .100

2) 5e^x = e^3x+1
.

We have: .e^3x+1 .= .5e^x

Divide by e^x: . e^2x+1 .= .5

Then: .2x + 1 .= .ln(5)

. . . . . . . . .2x .= .ln(5) - 1

. . . . . . . . . . . . . . .ln(5) - 1
. . . . . . . . . x . = . ----------- . ≈ . 0.3047
. . . . . . . . . . . . . . . . . 2