Proving a logarithmic equation

[Guest]

Hello, I'm new here and I was wondering if I could get some help with a problem. I have to prove the following:

b ^ log_b (x) = x.

Where ^ means exponent and _ means subscript.

My physics techer showed me one way to solve this, but on another forum(and some other research) I found that there's another way, but I don't understand how to do it.

Someone from another forum said b ^ log_b (x) = x precisely means log_b (x) and b ^ x are inverses. I know that those are inverses, but I want to know how you use this concept to proove that b ^ log_b (x) = x.

This has been very frustrating for me, and I would greatly appreciate help from someone.

 

[skeeter]

\(\displaystyle b^{log_b(x)} = x\)

take the log, base b, of both sides ...

\(\displaystyle log_b[b^{log_b(x)}] = log_b(x)\)

using the power rule for logs ...

\(\displaystyle log_b(x)log_b(b) = log_b(x)\)

since \(\displaystyle lob_b(b) = 1\) ...

\(\displaystyle log_b(x)*1 = log_b(x)\)

 

[Guest]

Ok thanks for the help, I now know another method to solve this, but I still want to know how to solve it using the mentioned method.

 

[stapel]

I have to prove the following: b^{log_b (x)} = x, where "_" means subscript.

Someone from another forum said b^{log_b (x)} = x precisely means log_b (x) and b^x are inverses. I know that those are inverses, but I want to know how you use this concept...

Hint: If f(x) and g(x) are inverses, then what is always the value of f(g(x)) and g(f(x))? :wink:

Eliz.

 

[Guest]

The value is always 1 if I'm not mistaken, and I know this, but how can you use this to prove the equation? That's what I'm really asking. Basically, here's what I need to know: f(x) and g(x) are inverse, multiplying them gets you a result of 1 I think. How does this apply in this case with exponents? A step by step method would be greatly appreciated, or at least an explanation of how you use this relationship in the equation.

 

[stapel]

The value is always 1....

Really? Try the following:

. . . . .f(x) = 3x + 4
. . . . .g(x) = (x - 4)/3

. . . . .f(x) = x², x > 0
. . . . .g(x) = sqrt[x]

. . . . .f(x) = ln(x), x > 0
. . . . .g(x) = e^x

What do you get when you compose the functions?

Eliz.

 

[Guest]

Sorry about that, it's x right? I meant multiplying inverses would get you the result 1, I didn't really look at how the question was phrased...

But could someone please show me how to proove the equation with this method?

I know multplying inverses results in 1, and results in x if multiplied by x...

1 * x = x

But this is confusing because of the fact that log is in the exponent... someone please show me how or at least show the first step for now? Basically I'm assuming you're saying that b ^ log_b (X) can be written as f(g(x)) or g(f(x)), but how do you get from b ^ log_b (X) to f(g(x))?

 

[Guest]

Someone please help. How does \(\displaystyle b^log_bx\) simplify to x?

 

[stapel]

Sorry about that, it's x right?

Composing inverse functions results in "x", yes.

I meant multiplying inverses would get you the result 1

Multiplying functions which are reciprocals would indeed result in "1". But reciprocals are very different from inverses, and multiplication is very different from composition.

Are you not familiar with function notation or function composition? Because an understanding of the concepts and techniques is necessary in order to apply the method in question.

Thank you.

Eliz.

 

[Guest]

Err, I'm not really familiar with those, could you clear it up for me and then show how it is used to solve the problem?

 

[stapel]

I'm not really familiar with those, could you clear it up for me...?

Since we cannot teach lessons here, please consider the following online resources instead:

. . . . .Functions: Concepts and Terms

. . . . .Function Notation and Evaluation

. . . . .Composition of Functions

. . . . .Inverse Functions

Once you have learned the basic concepts involved (assuming that you are familiar with exponentials and logarithms), then the result will be immediate. Just do the composition that they suggested, and apply the fact that the functions are inverses.

Eliz.

 

[Guest]

Err, ok then that's unfortunate, buI'll try it guess...

Are those lessons hard?

 

[Guest]

I took a look at them, helpful somewhat... but could you show the work anyway? Please? I'm still having a bit of trouble understanding how to apply this to the question...

 

[stapel]

...could you show the work anyway?

There is no "work" to do. The result is immediate. It follows directly from the definition of "inverse functions".

Eliz.

 

[Guest]

Ugh, I still don't get it. I really have no idea what to think of this. Isn't there anything you can say to help me figure it out?

 

[Guest]

Here's all I can think of:

g(f(x)) = f(g(x)) = x

f(x) and g(x) are inverses therefore we can let f(x) = log_b (x) and g(x) = b ^ x

Is this right?

 

[Guest]

Ah! I think I've got it now, my last post was right I believe.

g(f(x)) = f(g(x)) = x

g(x) = b ^ x
f(x) = log_b(x)

by subbing all the x's with f(x)'s we get:

g(f(x)) = b ^ f(x) = x

but f(x) = log_b(x)
therefore g(f(x)) = b ^ log_b(x) = x

There, I think I got it

 

[stapel]

Here's all I can think of:

g(f(x)) = f(g(x)) = x

f(x) and g(x) are inverses therefore we can let f(x) = log_b (x) and g(x) = b ^ x

Is this right?

That's it, exactly!

Eliz.