solve log10^(x+4) + log10 ^(x-2) = 1

I got x= -1 + sq root of 19 or -1 + sq root of 19. Just wanted to make sure I'm doing these right! thanks so much!

Question

solve log10^(x+4) + log10 ^(x-2) = 1

I got x= -1 + sq root of 19 or -1 + sq root of 19. Just wanted to make sure I'm doing these right! thanks so much!

lgbasallote · Answer

how does this look like?

$$\LARGE \log 10^{x+4} + \log 10^{x-2} = 1$$

lgbasallote · Answer

or is it $$\LARGE \log_{10} (x+4) + \log_{10} (x-2) = 1$$

anonymous · Answer

Yes, like the second equation, log base 10 :)

anonymous · Answer

$$log_{10}((x+4)(x-2))=1$$

anonymous · Answer

then use your log properties to change into exponential form

lgbasallote · Answer

i see..first you incorporate the product rule then
$$\log_{10} (x+4)(x-2) = 1$$
then you transform it into exponential form
$$10^1 = (x+4)(x-2)$$
$$10 = (x+4)(x-2)$$

lgbasallote · Answer

then you expand the right side

anonymous · Answer

dont forgot the domain x+4>0  and x-2>0  so  ur domain is x>-4

anonymous · Answer

ah I see thanks!!, so would you put the answer using the quadratic formula??

lgbasallote · Answer

i think it's non-factorable so yeah

anonymous · Answer

Great thank you!!

lgbasallote · Answer

anonymous · Answer

x= -1 - sq root of 19 and -1 + sq root of 19 but x=-1- sq root of 19 is not in domain so only answer is x=-1 + sq root of 19

anonymous · Answer

haha thank you both of you!! I would def give you both tips, however, I'm an unemployed pre-med student for now :)

anonymous · Answer

welcome :)