Solve: log x + log (x + 9) = 1?

Question

Please help -

Solve: log x + log (x + 9) = 1


1. 0
2. 1
3. 2
4. 3

Answer 1

Answer :2

log (1) + log (1+ 9) =1
0 +log10=1
0+1=1

Answer 2

Just remember that when you add two logs with the same base, you end up with a combination with multiplication, logx +log(x+9) ends up as just log x(x+9) [remember just log means base 10 and an ln means natural base(e)].
After you combine the logs you then need to put the new log into it's exponential form. The base stays and you switch the x and y terms, so,,,,,,,, log x(x+9) becomes 10^1=x(x+9). Distribute that out and move the ten over, factor and find the zeros. x^2 +9-10=0 becomes (x-1)(x+10)=0 x={-10,1} and since x can't be 0 or a negative term, 1 is your only possible solution.
Just remember that with logarithmic equations you need to isolate the log(combine the logs together to get one log on one side, just like we did up top) and have your y-term on the other side(the 1 in your problem), then just switch the x and y terms which then makes the log it's equivalent exponential form.
With exponential equations(you may have not learned these yet but you soon will if your doing these logarithims) you isolate the base with it's x on one side(the exponential term) and then just take the natural base of both sides of the equation and solve for x. Hope I helped

Answer 3

log x + log (x + 9) = 1

log [x(x + 9)] = 1

x(x + 9) = 10^1

x^2 + 9 x - 10 = 0

(x +10)(x - 1) = 0

x = -10 or x = 1, but x cannot be negative.

So, from your choices, it is choice 2. 1

Answer 4

log x + log (x + 9) =1
they have the same base which is log so

x(x + 9) =1
x^2 + 9x = 1
x^2 + 9x - 1 = 0
then do the completing the square