logb A = logb B turns out to be A = B. Why?

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nosit

Junior Member
Joined
Aug 9, 2020
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51
Hello,

I would like to understand algebraically what happens for an equation with both logs with the same base to become A = B. Example:

log (2x + 1) = log 11

2x + 1 = 11

Thank you in advance!
 
nosit said:
I would like to understand algebraically what happens for an equation with both logs with the same base to become A = B. Example:
log (2x + 1) = log 11
2x + 1 = 11
Especially, why both sides have to have the same base?
Click to expand...
Most impotently the logarithm function is an injection (one-to-one) function.
Thus if \(\log_b(X)=\log_b(Y)\text{ then }X=Y\) .
 
pka said:
Most impotently the logarithm function is an injection (one-to-one) function.
Thus if \(\log_b(X)=\log_b(Y)\text{ then }X=Y\) .
Click to expand...
Actually, it's the exact opposite of "impotent" -- it's "important"! (This is why I proofread.)

To put it another way, you can exponentiate both sides (raise b to each power), which eliminates the logs because the log is the inverse function of the exponential. If two numbers have the same log, then they must be the same number, because both are the same power of the base.
 
The log function is always increasing meaning that you can't take the log of two different values and get the same answer.
 
nosit said:
Hello,

I would like to understand algebraically what happens for an equation with both logs with the same base to become A = B. Example:

log (2x + 1) = log 11

2x + 1 = 11

Thank you in advance!
Click to expand...
The same reason as in:

b + (2x+1) = b + 11 .... \(\displaystyle \to\).... (2x+1) = 11 ....... (here I am NOT saying that "addition" is equivalent to "ln")

b * (2x+1) = b * 11 .... \(\displaystyle \to\).... (2x+1) = 11 ....... (here I am NOT saying that "multiplication" is equivalent to "ln")
 
Dr.Peterson said:
Actually, it's the exact opposite of "impotent" -- it's "important"! (This is why I proofread.)

To put it another way, you can exponentiate both sides (raise b to each power), which eliminates the logs because the log is the inverse function of the exponential. If two numbers have the same log, then they must be the same number, because both are the same power of the base.
Click to expand...
But posts that are NOT proof read are often more fun to read!
 
[MATH]log_a(x) = log_a(y) \implies x = a^{log_a(y)}.[/MATH] That's definitional.

[MATH]log_a(x) = log_a(y) \implies y = a^{log_a(x)}.[/MATH] That's also definitional.

[MATH]\therefore y = a^{log_a(y)} \ \because \ log_a(x) = log_a(y)[/MATH] by hypothesis.

[MATH]x = a^{log_a(y)} = y \implies x = y.[/MATH]
Two thing equal to a third thing are themselves equal.
 
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