Logarithmic functions: If ln(4) = a and ln(5) = b, then...

[Tazman]

If I have In(4) = a and In(5) = b

Does this mean that :

In(20) = Ina+Inb or a+b
In(16) = a^2

Am I understanding this correctly?
Thx for any help.
Taz

 

[stapel]

If I have In(4) = a and In(5) = b, [d]oes this mean that:

In(20) = Ina+Inb or a+b
In(16) = a^2

Don't try to do too many steps at once or in your head, as this can easily lead to the sort of confusion you're exhibiting. You're on the right track; you just need to write things out a bit more clearly. :wink:

. . . . .ln(20) = ln(4 * 5) = ln(4) + ln(5) = a + b

...so your second answer was correct. And:

. . . . .ln(16) = ln(4[sup:vwlh0b68]2[/sup:vwlh0b68]) = 2ln(4) = ..?

...or:

. . . . .ln(16) = ln(4 * 4) = ln(4) + ln(4) = a + a = ..?

Complete the steps to confirm the correct answer.

Eliz.

 

[Tazman]

If I have In(4) = a and In(5) = b, [d]oes this mean that:

In(20) = Ina+Inb or a+b
In(16) = a^2

. . . . .ln(16) = ln(4[sup:1zjps6th]2[/sup:1zjps6th]) = 2ln(4) = ..?
...or:
. . . . .ln(16) = ln(4 * 4) = ln(4) + ln(4) = a + a = ..?

Eliz - Did i Just not go far enough for my answer when I stopped at a[sup:1zjps6th]2[/sup:1zjps6th]).
Would it then be 2a?

 

[o_O]

Edit: Sorry, mistakenly posted something incorrect.

 

[Tazman]

Thanks o_o for confirming.

Taz

 

[stapel]

Eliz - Did i Just not go far enough for my answer when I stopped at a[sup:31mg1c1r]2[/sup:31mg1c1r]).
Would it then be 2a?

Since a[sup:31mg1c1r]2[/sup:31mg1c1r] does not equal 2a (unless a = 0 or a = 2, and we have no justification for assuming this), then, no, your "a[sup:31mg1c1r]2[/sup:31mg1c1r]" answer, as demonstrated, was incorrect. :!:

You'll need to use the log rules they taught you to evaluate these expressions. :wink:

Yep. Since "a" is a logarithm, then 2a and a[sup:31mg1c1r]2[/sup:31mg1c1r] are equivalent:

\(\displaystyle ln16 = \underbrace{ln4^{2}}_{a^{2}} = 2ln4 = 2a\)

No. There is a big difference between the log of a square, such as "ln(4[sup:31mg1c1r]2[/sup:31mg1c1r]) = 2ln(4)", and the square of a log, such as "ln[sup:31mg1c1r]2[/sup:31mg1c1r](4) = [ln(4)][sup:31mg1c1r]2[/sup:31mg1c1r]". :shock:

This may easily be confirmed numerically:

. . . . .log[sub:31mg1c1r]4[/sub:31mg1c1r](4[sup:31mg1c1r]2[/sup:31mg1c1r]) = 2 log[sub:31mg1c1r]4[/sub:31mg1c1r](4) = 2 * 1 = 2

...but:

. . . . .[log[sub:31mg1c1r]4[/sub:31mg1c1r](4)][sup:31mg1c1r]2[/sup:31mg1c1r] = (1)[sup:31mg1c1r]2[/sup:31mg1c1r] = 1

At a guess, the ambiguous notation in earlier posts may have been throwing you off the scent...?

Eliz.

 

[o_O]

Ah, I apologize. Haphazardly read through the posts and posted an incorrect answer

My apologies