noobmaster19
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- Joined
- Sep 10, 2020
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(3)^(1/x) = 2. Find x?
Hint: Use log-functions(3)^(1/x) = 2. Find x?
Actual, this was a logarithm question.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.
You say: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.
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.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.
Thank you. Understood.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).