logarithm power rules

[letoatreides3508]

I'm really having a hard time understanding the logarithm power rules.

log_aXⁿ = nlog_aX means that if the argument of a logarithm contains an exponent, the logarithm should be rewritten with the exponent as a co-efficient in front of the logarithm. But in mymathlab

log₄(3x²) = 2log₄(3x) is presented as a true or false question

When I say it's true, I get the problem wrong. What am I missing here (helpful replies only please)

 

[Romsek]

I'm really having a hard time understanding the logarithm power rules.

log_aXⁿ = nlog_aX means that if the argument of a logarithm contains an exponent, the logarithm should be rewritten with the exponent as a co-efficient in front of the logarithm. But in mymathlab

log₄(3x²) = 2log₄(3x) is presented as a true or false question

When I say it's true, I get the problem wrong. What am I missing here (helpful replies only please)

the problem is that only the x is squared not the 3.

what is true is that

log₄(3x²) = log₄((sqrt(3)x)²) = 2 log₄(sqrt(3)x)

 

[letoatreides3508]

Thanks, and just to verify

If the exponent is in the parantheses of the argument, it doesn't apply because the entire term isn't being squared, just the x. Am I understanding this?

 

[Romsek]

If the exponent is in the parantheses of the argument, it doesn't apply because the entire term isn't being squared, just the x. Am I understanding this?

I'm sorry I don't understand what you mean here.

The key is that if you're going to bring the 2 out then entire argument to the log function needs to be squared.

 

[letoatreides3508]

I mean

when do I put the co-efficient of the logarithm in front of log (as in nlog). And I took your answer to mean that only when the exponent is not in paranthesis in the argument.

 

[Romsek]

when do I put the co-efficient of the logarithm in front of log (as in nlog). And I took your answer to mean that only when the exponent is not in paranthesis in the argument.

log_aXⁿ = nlog_aX

It's all right here.

In your example the argument X to the log function is 3x²

But the equation above requires that the entire argument to the log function be raised to the n, in this case 2

So rewrite 3x² as (sqrt(3)x)² and now the entire argument to the log function is squared as required.

 

[stapel]

If the exponent is in the parantheses of the argument, it doesn't apply because the entire term isn't being squared, just the x. Am I understanding this?

Any portion of the argument of the log is going to be "in the parentheses of the argument", because that's where the argument goes: inside the parentheses. The point is that the power only comes out front if the power is on the entire argument. If the argument had been ((3x)^2), then the 2 could have come out. But since the argument was only (3x^2), the 2 couldn't come out.

Instead, you first have to apply another log rule to split the log: log(3x^2) = log(3) + log(x^2). And so forth.