I am solving this question:
$log_3(m-7)^2 = 4$
There are two ways to solve it.
The first way (expand the brackets):
$log_3(m^2 -14m + 49) = 4$
$m^2 – 14m + 49 = 3^4$
$m^2 – 14m – 32 = 0$
$m = 16,-2$
And both of these are valid solutions, if substituted into the original equation.
The second way (using logarithm power rule):
$2log_3(m-7) = 4$
$log_3(m-7) = 2$
$m-7 = 3^2$
$m-7 = 9$
$m = 16$
Which provides one solution, but misses the other solution.
My question is, where does the second solution fail, and why?
I have played around with these graphs in Desmos, and expect the problem lies the moment I apply the power rule. (Desmos only draws one branch of the logarithm function once I apply the power rule). I've discovered that I can solve this by writing $2log_3(|m-7|) = 4$ (with an absolute value) and I suspect that might be what I am supposed to do, but cannot for the life of me understand why, and this is not the way I was taught the logarithmic power rule in highschool.
Best Answer
The second solution failed because you should have written $\log_3(m-7)^2 = 2\log_3 |m-7|$. Be sure to always have positive argument in the log!