Evaluating Logarithmic equations

[SourPatchParent]

log₂16√8

I am having difficulty getting an answer that I am 100% sure about. Should I look at this like log₂16 *√8 first of all?

if so..

log₂16(8^½)

log₂₆₄
2^x=64
x=6 ??

 

[ksdhart]

What I'm confused about is how to jumped from the problem as written, with log₂(16) to log₂(64). My guess is you evaluated sqrt(8) and got 4, then multiplied that by 16. But sqrt(8) is not 4. You want to find a number such that x*x = 8. And 4 times 4 is 16, not 8.

As for the interpretation, without any parentheses, the problem is ambiguous. It could be interpreted as asking you to solve: log₂(16) * sqrt(8). In that case, you'd evaluate the log first, and then multiply by the square root of 8.

But, the problem could also be interpreted as asking: log₂(16*sqrt(8)). This seems to be what you solved, where you'd evaluate 16 times the square root of 8 first, and then take the log.

 

[stapel]

log₂16√8

I am having difficulty getting an answer that I am 100% sure about. Should I look at this like log₂16 *√8 first of all?

Probably not. Usually, for clarity's sake, multipliers go in front. Had the square root been meant to be multiplied against the log, it more likely would have been formatted as:

. . . . .\(\displaystyle \sqrt{8\,}\, \log_2(16)\)

Since the square root is appended after the log, I suspect that it is meant to be included in the argument of the log. (This, by the way, is why proper formatting uses parentheses to demark the argument. Shame on the assignment author for being so sloppy.) :shock:

So I suspect that the expression is meant to be as follows:

. . . . .\(\displaystyle \log_2\left(16\, \sqrt{8\,}\right)\)

...with the argument being able to be restated as:

. . . . .\(\displaystyle 16\, \sqrt{8\,}\, =\, \left(2^4\right)\, \left(\left(2^3\right)^{1/2}\right)\, =\, 2^4\, \cdot\, 2^{3/2}\)

log₂16(8^½)

log₂64
2^x=64
x=6 ??

Where did the "x" come from? Was the original expression meant to be an equation, so we need you to post the "equals" sign and whatever else was included in the original exercise? Or was the exercise really just the original expression? In either case, what were the instructions? If an equation, then you were probably supposed to "solve"; if the posted expression, then you were probably supposed to "expand" or "evaluate".

Please be complete. Thank you!

 

[SourPatchParent]

Probably not. Usually, for clarity's sake, multipliers go in front. Had the square root been meant to be multiplied against the log, it more likely would have been formatted as:

. . . . .\(\displaystyle \sqrt{8\,}\, \log_2(16)\)

Since the square root is appended after the log, I suspect that it is meant to be included in the argument of the log. (This, by the way, is why proper formatting uses parentheses to demark the argument. Shame on the assignment author for being so sloppy.) :shock:

So I suspect that the expression is meant to be as follows:

. . . . .\(\displaystyle \log_2\left(16\, \sqrt{8\,}\right)\)

...with the argument being able to be restated as:

. . . . .\(\displaystyle 16\, \sqrt{8\,}\, =\, \left(2^4\right)\, \left(\left(2^3\right)^{1/2}\right)\, =\, 2^4\, \cdot\, 2^{3/2}\)


Where did the "x" come from? Was the original expression meant to be an equation, so we need you to post the "equals" sign and whatever else was included in the original exercise? Or was the exercise really just the original expression? In either case, what were the instructions? If an equation, then you were probably supposed to "solve"; if the posted expression, then you were probably supposed to "expand" or "evaluate".

Please be complete. Thank you!

The question asks me to solve the equation. To be honest I included exactly what the original exercise said I have a question.. When you multiplied 2^3 * 2^(1/2) did you get 2^(3/2) because when you multiply exponents that are fractions you follow the multiplication rules for fractions? so it was 3/1* 1/2?

I use "x" because--following the trend with the whole lesson and the courses examples--I'm solving for the exponent that a power is multiplied by to get the sum in the logarithmic equation. In other words I believe I have to solve for the exponent on the base of the logarithm that equals .\(\displaystyle \left(16\, \sqrt{8\,}\right)\)

p.s I'm doing Advanced Functions through correspondence and that is one thing I've been noticing; the instructions to the "assignments" have not been very concise or clear at all. I would get a tutor if I could afford one..

 

[Deleted member 4993]

log₂16√8

I am having difficulty getting an answer that I am 100% sure about. Should I look at this like log₂16 *√8 first of all?

if so..

log₂16(8^½)

log₂₆₄
2^x=64
x=6 ??

I'll provide the answer now - since ~24 hrs. elapsed since the original post went up.

log₂[16(8^½)]

= log₂[2⁴(2^{3/2})]

= log₂[2^(4+3/2)]

= log₂[2^(11/2)] ← we know log_a[aⁿ] = n

= 11/2

= 5½

 

[Deleted member 4993]

2^x = 16SQRT(8)
x = LOG[16SQRT(8)] / LOG(2)
x = 5.5

Anything wrong with that, Sir Khan?

Nothing - except your way needs to use a calculator or computer.

My way - you would only need your head (the part of the body you use to play soccer - not hockey).

 

[lookagain]

log₂(16√8) =

log₂(16) + log₂(√8) =

log₂(16) + log₂[8^(1/2)] =

log₂(2^4) + log₂[(2^3)^(1/2)] =

log₂(2^4) + log₂[(2^(3/2)] =

4 + 3/2 =

8/2 + 3/2 =

11/2 =

5.5


.

 

[Steven G]

Nothing - except your way needs to use a calculator or computer.

My way - you would only need your head (the part of the body you use to play soccer - not hockey).

I used my head many times while playing hockey. As a goaltender I remember taking a number of shots to my face. Fortunately I was wearing my face mask. Before taking those shots to my head I was excellent at math, now i am so so.

 

[Otis]

The question asks me to solve the equation. To be honest I included exactly what the original exercise said...

... [I am] following the trend with the whole lesson and the courses examples

You left out the trend from the whole lessson, in your original post. First, write x as the value of the logarithm.

x = log₂[16*sqrt(8)]

Now you have an equation, and the instruction above makes sense.

When you multiplied 2^3 * 2^(1/2) did you get 2^(3/2) because when you multiply exponents that are fractions you follow the multiplication rules for fractions?

We always use the rule for multiplying fractions, when multiplying fractions, regardless of the position of the fractions (i.e., exponents or not).

Also, you have misinterpreted the expression (2^3)^(1/2). It doesn't represent 2^3 * 2^(1/2).

(2^3)^(1/2) = 2^(3*1/2)

I'm solving for the exponent that a power is multiplied by to get the sum in the [logarithm].

I think you misspoke.

A power is not multiplied by its exponent. The exponent is a part of the power.

A power: 2^x

2 is the base of the power; x is the exponent of the power. They are not multiplied.

Also, the logarithm's input is not a sum; it is a product. (16 times the square root of 8)

Hope this makes sense. Please ask, if you have questions about the terminology.

Cheers :cool:

 

[Otis]

2^x = 16SQRT(8)
x = LOG[16SQRT(8)] / LOG(2)

Please don't shout at Subhotosh.

 

[Steven G]

I got in a few fist fights while playing hockey:
my problem was always trying to block the other guy's fist with my nose...

That's one reason I always wore a mask.