Logarithm Problem: Multiple solutions to log_8(x) + log_8(x+6) = log_8(5x + 12)

[koreamaniac101]

I am relearning logarithms, and as I was solving a practice problem:

. . .log₈(x) + log₈(x + 6) = log₈(5x + 12)

I found that x is equal to both 3 AND -4

I know that negative logs can exist, as taking the log (base-10) of a decimal produces a negative exponent, but the answer showed x = 3 to be the only answer.

besides plugging the x's back into the equation, is there a rule or method for determining where an answer is valid or not?

-Soohan Kim

 

[HallsofIvy]

You say "I know that negative logs can exist" but you don't know what that means! Yes, if you take the logarithm, to any base, of a (positive) number less than 1, then the result is negative. That is not the same as saying you can take the logarithm of a negative number! With \(\displaystyle y= log_a(x)\), y can be negative, x cannot. The range of the logarithm function includes the negative numbers, the domain does not.

 

[Steven G]

I am relearning logarithms, and as I was solving a practice problem:

. . .log₈(x) + log₈(x + 6) = log₈(5x + 12)

I found that x is equal to both 3 AND -4

I know that negative logs can exist, as taking the log (base-10) of a decimal produces a negative exponent, but the answer showed x = 3 to be the only answer.

besides plugging the x's back into the equation, is there a rule or method for determining where an answer is valid or not?

-Soohan Kim

All the arguments have to be positive (just not negative is not enough as that allows 0). That is, x>0, x+6>0 and 5x+12>0. This is the same as x>0, x>-6 and x> -12/5. This means x>0.

And no, you did not find that x is equal to both 3 AND -4. You found that the possible answer to x are 3 and -4

 

[Yale]

I am relearning logarithms, and as I was solving a practice problem:

. . .log₈(x) + log₈(x + 6) = log₈(5x + 12)

I found that x is equal to both 3 AND -4

I know that negative logs can exist, as taking the log (base-10) of a decimal produces a negative exponent, but the answer showed x = 3 to be the only answer.

besides plugging the x's back into the equation, is there a rule or method for determining where an answer is valid or not?

-Soohan Kim

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The general way is as follow:
log₈(X) + log₈​(x+6) = log₈​(5x +12)
x(x+6) = 5x + 12
x²+x-12 =0
X=3 or -4
ln(x) is not defined for x <= 0
x = 3