Evaluating/Solving Logarithmic expressions/equations

[wvcaudill2]

Homework Statement

Evaluate the expression "log base 8 of 16"

Homework Equations

Don't know of any

The Attempt at a Solution

I don't know where to begin. I have read the textbook chapter on logarithms three times over, and I am now more thoroughly confused than ever. My textbooks seems to be skipping steps and citing the "Def. of Logarithm" as the reasoning.

For the problem above, my textbook says to set the expression equal to x, and then magically gets 8^x=16. After this, it gets an answer of 4/3, without mentioning how. I assume it wants me to look at the problem and be able to see that 8^(4/3)=16, but I don't think I can do that with all problems. I am confused as to why you set the expression equal to x, and then how the expression is rearanged to get 8^x.

Can someone please walk me through this problem without skipping any steps?

 

[Mark44]

Homework Statement

Evaluate the expression "log base 8 of 16"

Homework Equations

Don't know of any

The Attempt at a Solution

I don't know where to begin. I have read the textbook chapter on logarithms three times over, and I am now more thoroughly confused than ever. My textbooks seems to be skipping steps and citing the "Def. of Logarithm" as the reasoning.

For the problem above, my textbook says to set the expression equal to x, and then magically gets 8^x=16.

This isn't magic. They are using the definition of the logarithm, which says:
x = log_ay is equivalent to y = a^x

(There are some restrictions on y and a that I have omitted.)

For your expression, you have x = log₈ 16, which is equivalent to 16 = 8^x. There are no missing steps; the book is using the defining relationship between a log equation and the equivalent exponential equation.

It's useful to think of the log, base a, of a number as the exponent on a that produces that number. So if x is the log, base 8, of 16, then 8^x = 16.

Continuing the problem requires knowledge of the laws of exponents, one of which is that (a^m)ⁿ = a^mn. 16 is a power of 2, and so is 8. Another property is that a^m = aⁿ ==> m = n.

You have 16 = 8^x, so 2⁴ = (2³)^x.

Can you finish the problem now?

After this, it gets an answer of 4/3, without mentioning how. I assume it wants me to look at the problem and be able to see that 8^(4/3)=16, but I don't think I can do that with all problems. I am confused as to why you set the expression equal to x, and then how the expression is rearanged to get 8^x.

Can someone please walk me through this problem without skipping any steps?

 

[wvcaudill2]

Thanks, I see how to get 4/3 now.

What about this problem?

log base 6 of x = (.5 log base 6 of 9)+(.33 log base 6 of 27)

How do I get rid of the logs? Since they have the same base, can I just divide all the terms by log base 6 to cancel everything out?

 

[Mark44]

Thanks, I see how to get 4/3 now.

What about this problem?

log base 6 of x = (.5 log base 6 of 9)+(.33 log base 6 of 27)

How do I get rid of the logs? Since they have the same base, can I just divide all the terms by log base 6 to cancel everything out?

Absolutely not! log₆ is not a number -- it's a function

What properties of logs (any base) do you know? LIst the ones you know and we can go from there.

 

[wvcaudill2]

Well, I think this is the one applicable to this problem:

Product: log base b of mn = log base b of m + log base b of n

What do I do with the numbers in front of the logs though?

 

[Mark44]

Another one is log_b aⁿ = n log_ba

 

[wvcaudill2]

Ok, that's what I needed to solve this problem. Thanks!

 

[ibnsos]

Remember not to exponentiate unless you have only one log on one or both sides of the equal sign. Dunno why, but I always used to forget about that...

 

[Office_Shredder]

You can exponentiate whenever you want as long as you keep all the properties of exponentiation and logarithms straight

 

[The Chaz]

You can exponentiate whenever you want as long as you keep all the properties of exponentiation and logarithms straight

The difficulty in this prompts many to advise against exponentiating until exactly one log expression is on each side of the equality. But yeah, you can...

 

[paulfr]

Another solution is to raise both sides to a power of 6.

6 ^ (log base 6 of x) = 6 ^ [(.5 log base 6 of 9)+(.33 log base 6 of 27)]
Since 6^x and log base 6 of x are inverse functions, then
x = [ 6 ^ (.5 log base 6 of 9) ] [ 6 ^ (.33 log base 6 of 27) ]

Using the Power Rule for Logs ..

x = [ 9 ^ 0.5 ] [ 27 ^ 0.33 ]
x = 3 x 3
x = 9