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Sagot :
Given:
The equation is:
[tex]\ln e^{\ln x}+\ln e^{\ln x^2}=2\ln 8[/tex]
To find:
The solution for the given equation.
Solution:
We have,
[tex]\ln e^{\ln x}+\ln e^{\ln x^2}=2\ln 8[/tex]
It can be written as:
[tex]\ln x+\ln x^2=2\ln 8[/tex] [tex][\because \ln e^x=x][/tex]
[tex]\ln (x\cdot x^2)=2\ln 8[/tex] [tex][\because \ln a+\ln b=\ln (ab)][/tex]
[tex]\ln (x^3)=\ln 8^2[/tex] [tex][\because \ln x^n=n\ln x ][/tex]
On comparing both sides, we get
[tex]x^3=8^2[/tex]
[tex]x^3=64[/tex]
Taking cube root, we get
[tex]x=\sqrt[3]{64}[/tex]
[tex]x=4[/tex]
Therefore, the required solution is [tex]x=4[/tex].
Answer:
x=4
Step-by-step 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
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