Step-by-step explanation:

Solution :-

Given equation is 3^x + 3^(x+1) + 3^(x+2) =13

=> 3^x + [(3^x)×3] + [(3^x) ×3²] = 13

Since, a^m × a^n = a^(m+n)

=> 3^x + 3× 3^x + 9 × 3^x = 13

=> 3^x ( 1+3+9) = 13

=> 3^x (13) = 13

=> 3^x = 13/13

=> 3^x = 1

It can be written as

=> 3^x = 3⁰

We know that

If bases are equal then exponents must be equal.

Therefore, x = 0

Check :-

If x = 0 then LHS of the given equation is

3^x + 3^(x+1) + 3^(x+2)

=> 3⁰+(3⁰+¹) + (3⁰+²)

=> 3⁰+3¹+3²

Since, a⁰ = 1

=> 1+3+9

=> 13

=> LHS = RHS is true for x = 0

Verified the given relations in the given problem.

GɪᴠᴇN :-

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: 3 {}^{x} + 3 {}^{x + 1} + 3 {}^{x + 2} = 13 } \\ \end{gathered}[/tex]

Iᴅᴇɴᴛɪᴛʏ UsᴇD :-

[tex] \: \: \begin{gathered}\\ \large \bigstar \: \: \underline{ \boxed{\mathsf { \: \:x {}^{m} \times x {}^{n} = x {}^{(m + n)} \: \: }}} \: \: \bigstar \: \\ \end{gathered}[/tex]

Sᴏʟᴜᴛɪᴏɴ :-

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: 3 {}^{x} + 3 \times 3 {}^{x} + 9 \times 3 {}^{x} = 13} \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: 3 {}^{x} \: (1 + 3 + 9 ) = 13 } \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: 3 {}^{x} \: (13) = 13} \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: 3 {}^{x} = \frac{13}{13} } \\ \end{gathered}[/tex]

[tex] \begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: 3 {}^{x} = 1 } \\ \end{gathered}[/tex]

♪ We can compare or write this equation as,

[tex] \: \: \begin{gathered}\\ \large \bigstar \: \: \underline{ \boxed{\mathbf { \: \: x {}^{0} = 1 \: \: }}} \: \bigstar\\ \end{gathered}[/tex]

[tex]\begin{gathered}\\ \large\dashrightarrow\mathsf { \: \: 3 \red{{}^{x}} = 3 {}^ \red{0} } \\ \end{gathered}[/tex]

Hᴇɴᴄᴇ,

[tex]\begin{gathered}\\ \large\dashrightarrow \: \: \underline{ \boxed{\mathfrak \color{blue} { The \: Value \: of \: x = 0 \: \: } }}\\ \end{gathered}[/tex]

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★ Additional Information ★

[tex]\begin{gathered}\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \small \color{blue}{ \underline{\boxed{ \begin{array}{cc} \small \underline{\underline{\bf{ \color{red}{{ \orange \bigstar \: MᴏʀE \: IᴅᴇɴᴛɪᴛɪᴇS \: \orange \bigstar}}}}} \\ \\ \: \frak{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \: \frak{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \: \frak{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \: \frak{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \: \frak{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \: \frak{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \: \frak{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \: \frak{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \\ \: \end{array} }}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \: \: [/tex]

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