Log problem

[Baron]

Evaluate: n^(log (base a ) 2 + log (base a)64 )

My work.
n^(log (base a) 128)
n^(log (base a) 2^7)
n^(7log (base a) 2)

Then I'm stuck. Is this already simplified?

 

[mmm4444bot]

Evaluate: n^(log (base a ) 2 + log (base a)64 )

We cannot evaluate this expression without knowing the value of symbols n and a.

n^(log (base a) 128)
n^(log (base a) 2^7)
n^(7log (base a) 2)

Is this already simplified?

All three lines are simplifications of the given expression.

What exactly did they ask you to do ?

PS: We can use the underscore character to denote logarithm bases.

n^[log_a(2) + log_a(64)] = n^[7 log_a(2)]

 

[Baron]

Exact question was: Use the property that a^[log_a (x)] = x to evaluate n^[log_a (2) + log_a (64) ]

 

[mmm4444bot]

Ah, I never considered that n and a could be the same number.

Maybe there is a typographical error in your materials. I mean, perhaps they intended to type 'a' instead of 'n'.

Then, the exercise would be this:

Evaluate: a^[log_a(2) + log_a(64)]

And (using the referenced property), the evaluation would be this:

a^[log_a(128)] = 128

That's my best guess. Otherwise, there is NO WAY to evaluate the expression as given because we have no information about the relationship between the two numbers n and a.

I'm more interested in whether or not you understand the property.

The expression log_a(number) represents an exponent.

I mean, that's what logarithms are, exponents.

Sometimes, we don't know what an exponent is, so we need to express it using a logarithm.

For example, what exponent do we need on base 2 in order for the power to equal 5:

2^? = 5

We don't know (right away) what that exponent is, but we can at least write it as log_2(5).

So, log_a(number) is an exponent.

Specifically, it is the exact exponent needed on 'a' in order for the power to equal 'number'.

In other words, it's the exponent such that:

a^exponent = number

Therefore, if we do that (I mean, if we raise base a to that particular exponent), we get the power:

a^[log_a(number)]

and -- by the definition of logartihms -- this power obviously evaluates to 'number'.

Just like 2^[log_2(5)] = 5.


PS: Is your "teacher" a machine ? I find on-line courses to be rife with all sorts of mistakes.

 

[lookagain]

Post Deleted

 

[lookagain]

. . .

Baron,

a way among more ways to type a log expression is

"tex" inside of square brackets, a back slash, "log," an underscore, an open brace, the base, a close brace, the number
of which you're taking the logarithm, and a back slash followed by "tex," with both of these inside of square brackets.


Examples of where it might be used:

\(\displaystyle \log_{2}8\)


\(\displaystyle \log_{2}8 = 3\)


\(\displaystyle 2^{\log_{2}8} = 8\)


If you look at my text under the "QUOTE" button at the top right of this page, you should be able to see my code that I used.