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

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

2) log_{5}(2x-1) = 2

3) mx+1 : mx-1

Are these your intended questions Paulina? Number 3 doesn't make sense.

Paulina, you have to hit the "post answer" button at the bottom.

then tell me if what I have written is correct.

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Note to other people.

(I am putting Paulina's questions here because she is new and it is all too confusing)

Melody Jun 16, 2014

#1**+5 **

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

Melody Jun 16, 2014

#3**+10 **

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

As for the second question, we have........ log_{5}(2x-1) = 2.........in exponential form, we have

5^{2} = 2x-1

25 = 2x -1 subtract 1 from both sides

26 = 2x divide by 2

13 = x

And there you go......

CPhill Jun 16, 2014

#4**0 **

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???

Melody Jun 16, 2014