Question

log5(x+3) + log5(x-4) = 2log5(x-1)

Done it 4 times, and keep getting -13.

How? :(

Answer 1

it can't be -13 because you cannot take the log of a negative number

Answer 2

Yeah, I figured.

I turned the left handside into log(AB) and on the right handside, I moved the 2 over to square the (x-1) the bases are the same so I can just equate, but the x squared(s) cancel! :(

Answer 3

start with
\[\log\left((x+3)(x-4)\right)=\log\left((x-1)^2\right)\]

Answer 4

(x+3)(x-4)=(x-1)^2

Answer 5

ok then lets see
\[(x+3)(x-4)=(x-1)^2\]
\[x^2-x-12=x^2-2x+1\]
\[-x-12=-2x-1\]
\[x-12=-1\]
\[x=11\] is what i get

Answer 6

Damn, looks like I made a sign error! Thank god for that! Thank you so much Satellite. :)

Answer 7

i am going to make a guess
my guess is that you wrote
\[(x-1)^2=x^2-1\] which is wrong
just a guess though

Answer 8

yw

Answer 9

I'd recommend checking the answer again. If \(x=11\) then it implies that \(14\cdot7=100\) but this is obviously false.

Answer 10

I'm getting \[(x+3)(x-4)=(x-1)^2\]\[x^2-x-12=x^2-2x+1\]\[-x-12=-2x+1\]\[x-12=1\]\[x=13\]

Answer 11

Yeah, kinggeorge, i got a negative 13, so it was a sign error, thank you for the response.

Is it possible to ask you something in here?

Answer 12

Sure.

Answer 13

Lol I think it is typing error

Answer 14

He mistaken the -2x+1 as -2x-1

Answer 15

btw KingGeorge is right~

Answer 16

I'm a biomedical engineer, doing technological maths.

I can do differentiation, integration of the utmost difficulty without hesitation but its these simpler equations I am trying to remember how to do (its been a while)

There was a question I was given, that I was wondering if you could help with:

Q. Suppose the area of a rectangle is 114.4m2 and the length of the rectangle is 14m longer that its width. Calculate the length and width of the rectangle.

If i get that, I'm ready for my exam :)

Answer 17

cwtan, an error from me?

Answer 18

@satellite73 's typing error

Answer 19

To find the area of that, you get an expression \[x(x+14)=114.4\]Where \(x\) is the width, and \(x+14\) is the length.

Answer 20

Yeah, I did it right, and that becomes a quadratic?

The 2 x values you get, you discard one and refill in for X in Width + Length?

Answer 21

You would discard the negative x-value, and the other x-value would be your width.

Answer 22

Fantastic! You've been a great help my friend. I guess its time I head over to the biology side of this site and help some people? :)

Answer 23

You're welcome.