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Question

Question: How do you solve \(3{e^x} - 2 = 0\)?...

How do you solve 3ex2=03{e^x} - 2 = 0?

Explanation

Solution

In this question, we are given an equation in terms of ee (The natural base of logarithm). We have been asked to find the value of xx. Start by shifting all the constants to the right-hand side of the equation. In order to simplify ee, you will have to take log on both sides. After that, use certain logarithmic properties to simplify the left-hand side of the equation. Hence, you will get your answer.

Formula used:
logmn=nlogm\log {m^n} = n\log m

Complete step by step answer:
We are being given an equation in terms of natural base of logarithm, i.e., ee. Let us see how we can find the value of xx .
3ex2=03{e^x} - 2 = 0 …. (given)
Step 1: In this step, we will send all the constants to the other side such that only ee remains on the left-hand side.
3ex=23{e^x} = 2
Shifting the variable of ex{e^x} ,
ex=23{e^x} = \dfrac{2}{3}
Step 2: Now, we will take logs on both the sides of the equation.
logex=log23\log {e^x} = \log \left| {\dfrac{2}{3}} \right|
Step 3: Under this step, we will apply the following property of logarithm:
logmn=nlogm\log {m^n} = n\log m
Applying on the left-hand side,
x.loge=log23x.\log e = \log \left| {\dfrac{2}{3}} \right|
Naturally, the base of log is e. Since no base has been taken here, we will assume it to be e only.
Now, we know thatlogaa=1{\log _a}a = 1 . Using this in the above equation, we will put logee=1{\log _e}e = 1.
x(1)=log23x\left( 1 \right) = \log \left| {\dfrac{2}{3}} \right|
On simplification we get,
x=log23\Rightarrow x = \log \left| {\dfrac{2}{3}} \right|
Hence, the value of xx is log23\log \left| {\dfrac{2}{3}} \right|.

Note: 1) Let us study a little more about this ee.
The number of e'e', is also known as Euler’s number. It is a mathematical constant and it has an approximate value of 2.718282.71828. It is also the base of natural logarithm. Also, e'e' is an irrational number, i.e., it cannot be represented in the form of pq\dfrac{p}{q} .
2) I wrote above the identity logaa=1{\log _a}a = 1. Let us see from where we derived this.
We have a base conversion formula –
 lognm=logamlogan\ {\log _n}m = \dfrac{{{{\log }_a}m}}{{{{\log }_a}n}}
Using this formula, we can derive the formula that we used.
 logaa=logaloga\ {\log _a}a = \dfrac{{\log a}}{{\log a}}
Now, we will cancel out these terms and we will get our answer as 11.
Hence, logaa=1{\log _a}a = 1.