+118677 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)
+118677 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
+129852 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......
+118677 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???
