Question

How do I solve and check this log expression: log(x+2)+log(x-1)=1? HELP!

Answer 1

the answer is 3 but how without using trial and error?

Answer 2

log a+ log b=log(ab) right?
log(x+2)+log(x-1)=1
log((x+2)(x-1))=1
log((x+2)(x-1))=1
Can you finish this?

Answer 3

see cause after that it's: log(x^2+x-2)=1 and then I'm stuck...

Answer 4

log((x+2)(x-1))=log(10)
(x+2)(x-1)=10

Answer 5

You can't just randomly multiply the opposite side by log10 imron...

Answer 6

You have to do the same on both sides.

Answer 7

No I didn't.
I changed 1 to log10 cz log10=1

Answer 8

log10 is 2.3 some odd numbers. Or is this Log(base10)0?

Answer 9

I interpreted this problem as log(base10)(x+2)(x-1)=1

Answer 10

where are you guys getting log10?

Answer 11

log is the same as log(base10).

Answer 12

yes. But it says "1" in the end?

Answer 13

@ash.vasvani yes, i think the same of this problem.
@milliex51 log10 comes from that 1.
log10=1 (assuming base 10)

Answer 14

log(base10) is = to 1, as it is the same as writing 10^0, which is 1.

Answer 15

So imron, would we not get x+2(x-1)=0?

Answer 16

No,

Answer 17

Do you understand and agree until this part:
log((x+2)(x-1))=1

Answer 18

yes

Answer 19

log b of x = y
<=> x = b^y

Answer 20

okay

Answer 21

or think 1 = log 10 of 10

Answer 22

Since both sides have base 10 :
-> (x+2)(x-1) = 10

Answer 23

@milliex51 You should be able to solve the quadratic equation :)

Answer 24

but i thought it would be: log(base10)1 = 0...

Answer 25

'cause then 10^0 = 1?

Answer 26

Since the right side is 1, we can take its advantage to plug whatever base is!

Answer 27

Think log, not exponent!

Answer 28

What's your result?

Answer 29

Don't forget to validate the results!