#Need help with solving equations involving logarithms

25 messages · Page 1 of 1 (latest)

green elbow
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The correct answer is E, and I'm struggling to see how I can get that answer. I got to (1-m)/m = 3^(2x) but I'm not sure where to go from there. Any help would be greatly appreciated, thank you.

upper scarabBOT
mild ibex
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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)

green elbow
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that's what I got, but that answer is different from option E

mild ibex
green elbow
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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?

mild ibex
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the left hand side goes from 1-m to -m after dividing by -1?

green elbow
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ohhh i see

mild ibex
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the higher level issue here is that dividing by -1 really doesn't help solve for m

green elbow
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divding 1-m by 1 is just 1-m still right?

mild ibex
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yep

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I recommend taking a step back.
if you had a problem like
1-m = 55*m
how would you solve for m?

green elbow
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divide by 55?

mild ibex
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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

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or similarly useful, you could subtract each side by 55m

green elbow
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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

mild ibex
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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

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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

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1 = 3^2x * m + m

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1 = (3^2x + 1) * m (factored off m)

green elbow
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ohhhh ok I understand now, I didn't even think about factoring the right side

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thanks for the help

mild ibex
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🥳

green elbow
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.solved