Solve for x: log(x+2) + log(x-1)=1
I have no idea how to get to the answer o.O
I know the answer is 3, but I can't reach that. I know to use the logarithmic properties to make it
log[(x+2)(x-1)]=1

and then just

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

after that, I got either the answers x=11,8 or x=12,11

What am I doing wrong?

Question

Solve for x: log(x+2) + log(x-1)=1
I have no idea how to get to the answer o.O
I know the answer is 3, but I can't reach that. I know to use the logarithmic properties to make it
log[(x+2)(x-1)]=1

and then just 

(x+2)(x-1)=10

after that, I got either the answers x=11,8 or x=12,11

What am I doing wrong?

anonymous · Answer

@SarahEZZMcK

amistre64 · Answer

you can start by baseing each side .. assuming log is a base 10 in your class

anonymous · Answer

Right, I would've explained if the base was any different :)

amistre64 · Answer

in higher classes, there is only one log,  the natural.

log(x+2) + log(x-1)=1

10^( log(x+2) + log(x-1)=1)

10^log(x+2)  10^log(x-1) = 10^1

(x+2)(x-1) = 10

amistre64 · Answer

its a quadratic now;  we can expand and run the quadratic formula

amistre64 · Answer

x^2 +x - 2 = 10

x^2 +x - 12 = 0

anonymous · Answer

Why could I not simply take it as x+2=10 and x-1=10?

anonymous · Answer

Is there a tell for when to break it up like that or use the formula?

amistre64 · Answer

becuase 10*10 is not equal to 10

anonymous · Answer

I don't understand how that relates...

amistre64 · Answer

(x+2)(x-1) = 10 ... but 10 has many real factors so this setup is not all that useful.

if we know when  (x+2)(x-1) - 10 = 0 its more useful

amistre64 · Answer

if (x+2) = 10, then x-1=1 ... and we would have x=2

2+2 is not 10

amistre64 · Answer

if we have x-1 = 10,  then x+2 = 1  and x=-2

but -2-1 is not 10

amistre64 · Answer

lol,  not x=-2 but same issue develops

amistre64 · Answer

the issue is:

10*___ = 10  only if ___=1

the usefulness of equating stuff to zero is that:

0*____ = 0  is true for whatever ____ is

anonymous · Answer

aaaaaah!! ok, that makes more sense then :) So then, for normal parabola questions and such, I can simply break it up like so, but for logarithmic questions, I have to use the formula? Is that correct?

amistre64 · Answer

for this particular case;  the log can be simplified to a parabola question in order to find solutions

anonymous · Answer

Thank you very much! :D

amistre64 · Answer

once you find the solutions, you will need to dbl chk that they make sense in the original setup.  since log(a) only has a real value of a is greater than 0

anonymous · Answer

then since the answers to the formula were 3 and -4, 3 is the only possible answer. Got it :)