Question

can you please help me 3^1+x + 3^1-x = 10​

Answer 1

Answer:

• hope it helps you to

Explanation:

✍️✍️✍️by:- me

Answer 2

$\texttt{\textsf{\large{\underline{Solution}:}}}$

We have to find out the value of x in the given equation.

Given,

$\tt\implies 3^{1+x}+3^{1-x}=10$

We can also write it as,

$\tt\implies 3^{1}\cdot 3^{x}+3^{1}\cdot 3^{-x}=10$

$\tt\implies 3^{1}\cdot 3^{x}+\dfrac{3^{1}}{3^{x}}=10$

Let us assume that,

$\tt\implies u=3^{x}$

Therefore, the equation becomes,

$\tt\implies 3u+\dfrac{3}{u}=10$

$\tt\implies \dfrac{3u^{2}+3}{u}=10$

$\tt\implies 3u^{2}+3=10u$

$\tt\implies 3u^{2}-10u+3=0$

By splitting the middle term, we get,

$\tt\implies 3u^{2}-9u-u+3=0$

$\tt\implies 3u(u-3)-1(u-3)=0$

$\tt\implies (3u-1)(u-3)=0$

Therefore,

$\begin{cases}\tt 3u-1=0\\ \tt u-3=0\end{cases}$

On Solving, we get,

$\tt\implies u=3,\dfrac{1}{3}$

When u = 3,

$\tt\implies 3^{x} = 3$

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

$\tt\implies x=1$

When u = 1/3,

$\tt\implies 3^{x} = \dfrac{1}{3}$

$\tt\implies 3^{x} = 3^{-1}$

$\tt\implies x=-1$

⊕ Therefore, the values of x are 1 and -1.

$\texttt{\textsf{\large{\underline{Answer}:}}}$

• x = 1, -1.