Question

if 3^x+3=1/(81) ^x-2 find the value of (3x)^2x​

Answer 1

Answer:

${3}^{x + 3} = \frac{1}{ {(81)}^{x - 2} } \\ \\ {3}^{x + 3} = \frac{1}{ {( {3}^{4} )}^{x - 2} } \\ \\ {3}^{x + 3} = \frac{1}{ {(3) }^{4x - 8} } \\ \\ {3}^{x + 3} \times {(3) }^{4x - 8} = {3}^{0} \\ \\ {3}^{x + 3 + 4x - 8} = {3}^{0} \\ {3}^{5x - 5} = {3}^{0} \\ \color{blue} \underline{on \: comparing}: \\ 5x - 5 = 0 \\ \boxed{x = 1} \\ \color{blue} \underline{put \: the \: value \: of \: x \: in \: {3x}^{2x} }: \\ {3x}^{2x} = {(3 \times 1)}^{2 \times 1} = {3}^{2} = 9 \\ \\\red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$

Answer 2

$\rm\Huge\underline{\text{Solution}}$

$\rm\large\underline{\text{Exponential equations}}$

→ The exponential equations have variables in the exponents. To solve the equations, usually, we use substitution, comparison of exponents.

$\rm\cdots\longrightarrow3^{x+3}=\dfrac{1}{81^{x-2}}$

$\rm\cdots\longrightarrow3^{x+3}=3^{-4(x-2)}$

$\rm\cdots\longrightarrow 3^{x+3}=3^{-4x+8}$

Both powers are equal if both the exponents are equal.

$\rm\cdots\longrightarrow x+3=-4x+8$

$\rm\cdots\longrightarrow5x=5$

$\rm\cdots\longrightarrow x=1$

Hence, $3^{2}=9$ is the answer.

$\rm\Huge\underline{\text{Learn more}}$

$\rm\large\underline{\text{Advanced questions}}$

Determine the number of the solutions of $\rm2^x+4^x=8^x+16^x$.

Taking substitution,

$\rm\cdots\longrightarrow\underline{t=2^x}$

The equation becomes a biquadratic equation,

$\rm\cdots\longrightarrow t+t^2=t^3+t^4$

$\rm\cdots\longrightarrow t^4+t^3-t^2-t=0$

$\rm\cdots\longrightarrow t(t^2-1)(t+1)=0$

$\rm\cdots\longrightarrow t=-1\text{(Double solution)}\text{, or }t=0\text{, or }t=1$

As $\rm2^{x}>0$ for real $\rm x$, $\rm 2^x=-1$ and $\rm2^x=0$ cannot have a solution.

Hence $\rm2^x=1$ or $x=0$, only one solution exists.