+0

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

+1
587
6
+95361

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.

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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)

Jun 16, 2014

#5
+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

Jun 16, 2014

#1
+95361
+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

Jun 16, 2014
#2
+95361
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Now Paulina - you have to respond by hitting the post answer button.

Jun 16, 2014
#3
+94607
+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......

Jun 16, 2014
#4
+95361
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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???

Jun 16, 2014
#5
+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
#6
+95361
0

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

Jun 17, 2014