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What is the true solution to the equation below?

l n e Superscript l n x Baseline + l n e Superscript l n x squared Baseline = 2 l n 8
x = 2

User Albert Schilling
by
2.8k points

2 Answers

23 votes
23 votes

Answer:

x=4

Explanation:

What is the true solution to the equation below?

ln e Superscript ln x Baseline + ln e Superscript ln x squared Baseline = 2 ln 8

x = 2

x = 4

x = 8

User Chaggster
by
3.3k points
12 votes
12 votes

Given:

The equation is:


\ln e^(\ln x)+\ln e^(\ln x^2)=2\ln 8

To find:

The solution for the given equation.

Solution:

We have,


\ln e^(\ln x)+\ln e^(\ln x^2)=2\ln 8

It can be written as:


\ln x+\ln x^2=2\ln 8
[\because \ln e^x=x]


\ln (x\cdot x^2)=2\ln 8
[\because \ln a+\ln b=\ln (ab)]


\ln (x^3)=\ln 8^2
[\because \ln x^n=n\ln x ]

On comparing both sides, we get


x^3=8^2


x^3=64

Taking cube root, we get


x=\sqrt[3]{64}


x=4

Therefore, the required solution is
x=4.

User Doody P
by
3.0k points