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Question: How do you simplify \({e^{2 + \ln x}}\)?...

How do you simplify e2+ln⁡x{e^{2 + \ln x}}?

Explanation

Solution

: In order to determine the value of the above question ,first separate the exponent into the product of two by using the exponent property ab+c=abac{a^{b + c}} = {a^b}{a^c}and then using the fact that the constant e raised to the power ln⁡x\ln xis imply equal to xxyou will get your required result the value of constant e=2.71828e = 2.71828

Complete step by step solution:
Applying the above exponent rule in this question, we get:
ab+c=abac{a^{b + c}} = {a^b}{a^c}
e2+ln⁡x{e^{2 + \ln x}}=eln⁡(x)e2{e^{\ln (x)}}{e^2}----eq1
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
But we also need to know that the number eeand log⁡x\log xare actually inverses of each other.
First, we are going to rewrite and solve eln⁡(x){e^{\ln (x)}} with the help of the following properties of natural logarithms.
Lety=eln⁡(x)y = {e^{\ln (x)}}
Applying the natural logarithm to both the sides
Also applying the logarithmic property
nlog⁡m=log⁡mn mlog⁡(n)=n  n\log m = \log {m^n} \\\ {m^{\log (n)}} = n \\\
ln⁡(y)=ln⁡eln⁡(x)\ln (y) = \ln {e^{\ln (x)}}
Note that ln(e)=1
ln⁡(y)=ln⁡(x).ln⁡(e)\ln (y) = \ln (x).\ln (e)
ln⁡(y)=ln⁡(x).1\ln (y) = \ln (x).1
ln⁡(y)=ln⁡(x)\ln (y) = \ln (x)
Logs with the same base are equal . therefore they are equal and we can write it in simplified manner as
y=xy = x
Also
y=eln⁡(x)y = {e^{\ln (x)}}
Therefore ,eln⁡(x)=x{e^{\ln (x)}} = x-----eq2
So, we can substitute the eq2 in eq1,
We can write it as e2+ln⁡x=eln⁡(x)e2 e2+ln⁡x=x.e2  {e^{2 + \ln x}} = {e^{\ln (x)}}{e^2} \\\ {e^{2 + \ln x}} = x.{e^2} \\\
Therefore ,simplification of e2+ln⁡x{e^{2 + \ln x}} is x.e2x.{e^2}.

Note: 1. Value of the constant ”e” is equal to 2.71828.
2. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.
3.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
log⁡b(mn)=log⁡b(m)+log⁡b(n){\log _b}(mn) = {\log _b}(m) + {\log _b}(n)
4. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
log⁡b(mn)=log⁡b(m)−log⁡b(n){\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)
5. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
nlog⁡m=log⁡mnn\log m = \log {m^n}
6. The above guidelines work just if the bases are the equivalent. For example, the expression log⁡d(m)+log⁡b(n){\log_d}(m) + {\log _b}(n)can't be improved, on the grounds that the bases (the "d" and the "b") are not the equivalent, similarly as x2 × y3 can'to be disentangled on the grounds that the bases (the x and y) are not the equivalent.
7. Don’t forget to cross check your result.