#Need help with solving equations involving logarithms
25 messages · Page 1 of 1 (latest)
yes, multiplying both sides by m is a good step. at this point, you can treat 3^2x as just some number (you can replace it with some variable if you prefer)
that's what I got, but that answer is different from option E
double check your work from the 2nd to the 3rd line
from diving both sides with 1 I got -m = 3^(2x) * m, then that third line is me dividing both sides by -1. i'm confused?
the left hand side goes from 1-m to -m after dividing by -1?
ohhh i see
the higher level issue here is that dividing by -1 really doesn't help solve for m
divding 1-m by 1 is just 1-m still right?
yep
I recommend taking a step back.
if you had a problem like
1-m = 55*m
how would you solve for m?
divide by 55?
well, when we're "solving" for m, we're really trying to move all the m values to one place
if you divide by 55, there's still an m on the left and right hand sides, so it isnt any better.
you could instead add m to both sides, and that removes the m on the left
or similarly useful, you could subtract each side by 55m
so i would do this next? I'm just not understanding what I'm supposed to be doing on the left, like how I can get m by itself
yeah, that's why it was better to have multiplied both sides by m (per your original question), which this newest work just undid
a fraction where m is in multiple places is complicated, so we want to stretch it out, bringing us to/returning us to:
1 - m = 3^2x * m
we see some terms have m as a factor, and some don't, so the next best step is to consolidate them, so we can factor m off
1 - m + m = 3^2x * m + m
1 = 3^2x * m + m
1 = (3^2x + 1) * m (factored off m)
ohhhh ok I understand now, I didn't even think about factoring the right side
thanks for the help
🥳
.solved