#Logarithmic Equation

33 messages · Page 1 of 1 (latest)

hybrid fable
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log(x, 8) = log(x, 2³) = 3 log(x, 2)
log(x², 16) = log(x², 2⁴) = (4/2) log(x, 2) = 2 log(x, 2)
3 log(x, 2) - 2 log(x, 2) = 1
log(x, 2) = 1
x¹ = 2
x = 2

solid lily
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...change of base formula. It's always change of base formula.

quartz ridge
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I had to use a more, "additional logarithm law" which isn't allowed supposedly.

solid lily
quartz ridge
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neither have I

solid lily
quartz ridge
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It's not the solution though?

solid lily
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It's the solution I got.

quartz ridge
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plug it in and check

solid lily
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...yeah, it works perfectly.

hybrid fable
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why? log(a^n, b^m) = (m/n) log(a, b)

quartz ridge
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with the domain restrictions

hybrid fable
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also if plugged in 2 the equation satisfies

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no

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while this one doesn't

solid lily
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log_a^x(b^y)
= ln(b^y)/ln(a^x)
= yln(b)/xln(a)
= y/x * ln(b)/ln(a)
= y/x log_a(b)```
quartz ridge
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Okay the way I did it was

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log _x 8 = log _x^2 8^2 = log_x^2 64
log_x^2 (64/16) = 1
x^2 = 4
x = + or minus 2

solid lily
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Change of base formula is basically invaluable for logarithm work.

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I did: log_x(8) - log_x^2(16) = 1 ln(8)/ln(x) - ln(16)/ln(x^2) = 1 3ln(2)/ln(x) - 4ln(2)/2ln(x) = 1 (3ln(2) - 2ln(2))/ln(x) = 1 ln(2) = ln(x) x = 2

quartz ridge
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both works though right

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the solution would be 2 with the domain restriction

solid lily
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Any time you have a sum or difference of logarithms with a different base, do change of base.

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...yes, because of the change of base formula.

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No, this is true because of the change of base formula.

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...okay, back up. What two things do you think I'm saying are called the change of base formula?

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It's a best practice.

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There's no reason to try all these funny little tricks if you just do CoB instead.

quartz ridge
solid lily
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That's because it's stupid and pointless when you could just use CoB.

solid lily
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You've never seen the change of base formula???