+0  
 
0
513
6
avatar+93629 

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

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

2) log5(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.

--------------------------------------------------

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

Best Answer 

 #5
avatar+2353 
+10

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

reinout-g  Jun 16, 2014
 #1
avatar+93629 
+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
 #2
avatar+93629 
0

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

Melody  Jun 16, 2014
 #3
avatar+89803 
+10

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......

CPhill  Jun 16, 2014
 #4
avatar+93629 
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
 #5
avatar+2353 
+10
Best Answer

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

reinout-g  Jun 16, 2014
 #6
avatar+93629 
0

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

 

Melody  Jun 17, 2014

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