How Do We Simplify and Interpret the Expression e^(ln(4)/2)?

In summary: So that's why the intermediate result 4 is there above, and that's why the square root of 4 is taken.In summary, the equation ln(x^2) = 4 can be rewritten as x = ± (e^2) or x = ± (sqrt(e^(ln(4)))). The intermediate result of 4 is present because it is the specific exponent that when used with e^1 results in 4. This is why the square root of 4 is taken in the final equation.
  • #1
late347
301
15

Homework Statement


$$ln(x^2)=4$$

Homework Equations


##a^{log_a(x)}=x##
##log_a(a^x)=x##

The Attempt at a Solution

ln(x^2)=4

<=> ##2ln(x)=ln(4)##

<=> ##ln(x)= [ln(4)]/2##

<=> ##log_e(x)= [ln(4)]/2##

<=> ##e^{ln(4)/2}=x##

<=> ##[e^{ln(4)}]^{1/2}##

<=> ##sqrt(e^{ln(4)})##

here I was a little bit confused about, how can we know what the thing inside the square root will be, in order to take the square root from it?
What do we know about the value of (e^{ln(4)}) so the square root of it can be taken?

I understand the other formula which was
##log_a(a^x)=x##
obviously the ##log_a(a^x)## asks us what exponent is the correct one, when you want to raise a to the exponent of something, such that the result will become a^x. The answer is x for the exponent.
 
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  • #2
oops I made a terrible mistake there
It helps to read with your brain and eyes. Instead of just eyes and not using the brain.

ln(x^2)=4

<==>
log_e(x^2) =4 {at this point I think we must keep the squared term instead of using dropping the exponent down such as log(x^r) = r*log(x)}
<==>
e^4= x^2
<==>
x= ± (e^2)
 
  • #3
late347 said:
oops I made a terrible mistake there
It helps to read with your brain and eyes. Instead of just eyes and not using the brain.

ln(x^2)=4

<==>
log_e(x^2) =4 {at this point I think we must keep the squared term instead of using dropping the exponent down such as log(x^r) = r*log(x)}
<==>
e^4= x^2
<==>
x= ± (e^2)

Alternatively, you could write ##|x| = e^2.##
 
  • #4
What about the log formula being used at another example such as

##e^{log_e(4)÷2}=?##

I have a harder time justifying that... I know that the result is 2 because I used a calculator.

You could arrange that such as
##sqrt(e^{ln(4)})##

Hmmm... now that I think about it more...

It seems that ln4 means... the exponent (lets call it with variable r ) with the requirement that it is the specific exponent which is required to be used to raise (e^1) such that the result of e^r=4

So therefore it seems that the intermediate result in the exponent spot there... e^{ln4} = 4
Because e^r=4.

And square root of 4 is 2
 

Related to How Do We Simplify and Interpret the Expression e^(ln(4)/2)?

1. What is the basic log formula?

The basic log formula is logb(x) = y, where x is the number being raised to a power, b is the base of the logarithm, and y is the power to which b must be raised to equal x.

2. How do I solve a logarithmic equation?

To solve a logarithmic equation, use the basic log formula to rewrite the equation in exponential form and then solve for the variable. For example, if you have log2(x) = 3, you can rewrite this as 23 = x, so x = 8.

3. What is the inverse of a logarithm?

The inverse of a logarithm is an exponential function. For example, if you have log2(x) = y, the inverse function would be 2y = x.

4. How do I graph a logarithmic function?

To graph a logarithmic function, first determine the domain and range of the function. Then, use a table of values to plot points and connect them with a smooth curve. Remember that the base of the logarithm will affect the shape of the graph.

5. What are some common uses of logarithms?

Logarithms are commonly used in scientific and mathematical calculations, such as in chemistry to measure pH levels and in finance to calculate compound interest. They are also used in computer science and engineering for data compression and signal processing.

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