Question

Find the value of $x$ in
$3^x(1+3^x)=2$
??✨​

Answer 1

Question

Find the value of a real number $x$ in the equation $3^{x}(1+3^{x})=2$.

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Solution

In this exponent equation, we can find common terms. Let the common term be $3^x=t$.

$\implies t(1+t)=2$

$\implies t^2+t-2=0$

$\implies (t+2)(t-1)=0$

This results to $t=-2$ or $t=1$. However, $t=-2$ cannot be valid because the graph of $y=3^{x}$ is always above the x-axis. (Attachment included.) This leads to $3^{x}>0$ or $t>0$.

Now $t=1$ is obtained by solving the quadratic equation.

$t=1$

$\implies 3^{x}=1$

$\implies \log_{3}3^{x}=\log_{3}1$

$\implies x\log_{3}3=0$

$\implies x=0$

So, the value of $x$ is $0$.

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This is the required answer.

Answer 2

Answer:

x= log(-1/2+√13/2 ) to the base 3

(or)

x=log(-1/2-√13/2 ) to the base 3

Step-by-step explanation:

let 3^x=a

then a (1+a)=3

a^2+a=3

a^2+a-3=0

by applying quadratic formula

a=-1/2+√13/2 (or) -1/2-√13/2

so 3^x=-1/2+√13/2

x= log(-1/2+√13/2 ) to the base 3

(or)

x=log(-1/2-√13/2 ) to the base 3

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