Can you help me with this exercises? please. ;D!

I'm not 100% sure but I thought there's some rule that

$$ln(x^k)\overset{for x>0,k>0}{=}k*ln(x)$$

I think that for x<0 it's simply not true

1) ln (4-x) + ln(2) = 2 ln(x)

$$\begin{array}{rll} ln(2(4-x))&=&ln(x^2)\\\\ 8-2x&=&x^2\\\\ x^2+2x-8&=&0\\\\ (x+4)(x-2)&=&0\\\\ x=-4\qquad &or& \qquad x=2 \end{array}$$

now I am confused as well.

You cannot have  the ln of a negative number so I guess x=2

Now Paulina - you have to respond by hitting the post answer button.

Melody's first answer is correct....here's a graph of the solution....

As for the second question, we have........ log₅(2x-1) = 2.........in exponential form, we have

5² = 2x-1

25 = 2x -1     subtract 1 from both sides

26 = 2x         divide by 2

13 = x

And there you go......

This one was confusing me a bit.

ln (x)² = 2ln(x)

but with 2ln(x) x must be >0

and with ln (x)²  x is just in the set of reals.

How does that work???

That is interesting reinout.  I guess it has to be something like that.  Thankyou for commenting on this one.