Solving this exponetial. Find x?

To solve any equation of the form f(x)= y you need to use the inverse function to f: \(\displaystyle x= f^{-1}(y)\). The inverse function to \(\displaystyle f(x)= a^x\) is \(\displaystyle f^{-1}(x)= log_a(x)\).

Start by taking a logarithm of both sides. The base of the logarithm is your choice.
 
HallsofIvy said:
To solve any equation of the form f(x)= y you need to use the inverse function to f: \(\displaystyle x= f^{-1}(y)\). The inverse function to \(\displaystyle f(x)= a^x\) is \(\displaystyle f^{-1}(x)= log_a(x)\).

Start by taking a logarithm of both sides. The base of the logarithm is your choice.
Click to expand...
Actual, this was a logarithm question.
The question was as follows :

log2(3) = x --->(2 is the base in the logarithm)

=> 2^x = 3
=> 2= (3)^(1/x)

I don't further how to proceed.

This is where i was stuck. I was trying to understand logarithms and I was trying to solve this without the logarithm table and calculator.
Sorry, I'm don't know how to use this text box for writing equations.
 
noobmaster19 said:
Actual, this was a logarithm question.
The question was as follows :

log2(3) = x --->(2 is the base in the logarithm)

=> 2^x = 3
=> 2= (3)^(1/x)

I don't further how to proceed.

This is where i was stuck. I was trying to understand logarithms and I was trying to solve this without the logarithm table and calculator.
Sorry, I'm don't know how to use this text box for writing equations.
Click to expand...
You say:

The question was as follows :​
log2(3) = x --->(2 is the base in the logarithm)​

But there is NO QUESTION above - what were you asked to find?
 
You are making no sense! Your initial post said "find x". But now you are saying
"The question was as follows :

log2(3) = x"

As Subhotosh Khan said, there is no question there!

And, as far as "finding x" is concerned, \(\displaystyle x= log_2(3)\)..........................edited
 
Last edited by a moderator:
noobmaster19 said:
Actual, this was a logarithm question.
The question was as follows :

log2(3) = x --->(2 is the base in the logarithm)

=> 2^x = 3
=> 2= (3)^(1/x)

I don't further how to proceed.

This is where i was stuck. I was trying to understand logarithms and I was trying to solve this without the logarithm table and calculator.
Sorry, I'm don't know how to use this text box for writing equations.
Click to expand...
I wonder if you are saying that you want to find a way to calculate [MATH]\log_2(3)[/MATH] without a table or a calculator, and you think that rearranging the equation as you did will help. It won't.

There is a reason you are expected to use a table or calculator! The table (many years ago) probably took years to make. The calculator does a lot of work (not just plugging into a formula) to get a result. There are ways to do the same work by hand, but knowing them wouldn't help you understand logarithms. One way would be just to try calculating [MATH]2^x[/MATH] for many values of x, adjusting bit by bit until the result is as close to 3 as you desire. Other methods are basically ways to speed that up.

Don't worry about writing equations more neatly; that feature is there mostly for the helpers to use. What you're doing is fine (except that log_2(3) would be clearer).
 
Dr.Peterson said:
I wonder if you are saying that you want to find a way to calculate [MATH]\log_2(3)[/MATH] without a table or a calculator, and you think that rearranging the equation as you did will help. It won't.

There is a reason you are expected to use a table or calculator! The table (many years ago) probably took years to make. The calculator does a lot of work (not just plugging into a formula) to get a result. There are ways to do the same work by hand, but knowing them wouldn't help you understand logarithms. One way would be just to try calculating [MATH]2^x[/MATH] for many values of x, adjusting bit by bit until the result is as close to 3 as you desire. Other methods are basically ways to speed that up.

Don't worry about writing equations more neatly; that feature is there mostly for the helpers to use. What you're doing is fine (except that log_2(3) would be clearer).
Click to expand...
Thank you. Understood.
 
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