- #1
Helly123
- 581
- 20
Homework Statement
Homework Equations
log_2 x = y
2^y = x
3^2^y
The Attempt at a Solution
log_2 x = y
2^y = x
log_2 {log _3 {log _2 { log_3 {2^y} } } }
what am I suppose to do?
all I know the ' a ' must be greater than zero.fresh_42 said:Try to work from outside in.
##f(a)= \log_2(a)## for some ##a##. Where is it defined, i.e. which values for ##a## are allowed? .
fresh_42 said:Try to work from outside in.
##f(a)= \log_2(a)## for some ##a##. Where is it defined, i.e. which values for ##a## are allowed? Say allowed the values are the set ##A_a##.
Next you have ##f(b) = \log_3(b)## where ##f(b) \in A_a##. What does this mean for ##b##? Say we get allowed values in ##A_b##.
Next you have ##f(c) = \log_2(c)## where ##f(c) \in A_b##. What does this mean for ##c##?
And so on, until ##f(x)=\log_2(\log_3(\log_2(\log_3(\log_2(x))))) = f(e)##. The set ##A_e## is the solution.
##f(x) = \log_2(a)##fresh_42 said:I meant what I wrote: outside in. (Maybe I had a step too many or a set ##A_*##.)
We have ##f(x) = \log_2(sth.)## at the start. I called that something ##a## which actually is ##a=\log_3(\log_2(\log_3(\log_2(x))))##, but I don't care by now. Then we have, as you've said, ##a > 0##. Now ##a = \log_3(b) > 0## for a new something called ##b##, which is ##b=\log_2(\log_3(\log_2(x)))##. Which ##b## are here allowed? Etc.
I meant something like this :fresh_42 said:You included it, as you said ##a>0##, which is the first condition. Here ##f(a)=\log_2(a)## has still all reals as possible values. But in order to have a number ##a>0## written as another logarithm, the next one ##a=\log_3(b)\; , \;b>0## isn't sufficient, because e.g. ##\log_3(\frac{1}{3}) = -1## which isn't positive. So ##b## has to be at least ##1##.
To solve a logarithm within a logarithm, you need to use the power rule of logarithms. This means that you can rewrite the logarithm as an exponent and then use the properties of exponents to simplify the expression.
The domain of a logarithmic function is all the values that the input (x) can take. In other words, it is all the possible values that you can plug into the logarithm and get a valid output. For most logarithmic functions, the domain is all positive real numbers.
To find the maximum domain of a logarithm within a logarithm, you need to look at the restrictions on the input of both logarithms. The maximum domain is the set of values that satisfy both restrictions. You may need to use algebraic manipulation to determine the maximum domain.
A logarithm is the inverse function of an exponential function. This means that a logarithm "undoes" the action of an exponential function. In other words, if an exponential function raises a base to a power, the logarithm with the same base "undoes" that action by finding the power that the base was raised to.
No, the input of a logarithmic function must always be a positive number. This is because the logarithm is the inverse of an exponential function, and an exponential function can only produce positive results. Therefore, the input of a logarithm cannot be negative.