Appropriate Question :- Solve
[tex]\sf \: 3^{x} + 3^{x + 1} + 3^{x + 2} = 13 \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given equation is
[tex]\sf \: 3^{x} + 3^{x + 1} + 3^{x + 2} = 13 \\ \\ [/tex]
We know,
[tex]\boxed{ \sf{ \: {x}^{m + n} = {x}^{m} \times {x}^{n} \: }} \\ \\ [/tex]
So, using this law of exponents, above equation can be rewritten as
[tex]\sf \: 3^{x} + 3^{x} \times {3}^{ 1} + 3^{x} \times {3}^{2} = 13 \\ \\ [/tex]
[tex]\sf \: 3^{x} (1+{3}^{ - 1}+{3}^{2}) = 13 \\ \\ [/tex]
[tex]\sf \: 3^{x} (1+3 + 9) = 13 \\ \\ [/tex]
[tex]\sf \: 3^{x} (13) = 13 \\ \\ [/tex]
[tex]\sf \: {3}^{x} = \dfrac{13}{13} \\ \\ [/tex]
[tex]\sf \: {3}^{x} = 1 \\ \\ [/tex]
[tex]\sf \: {3}^{x} = {3}^{0} \\ \\ [/tex]
[tex]\bf\implies \:x = 0 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
x = 0
Step-by-step explanation:
To Find:
Formula used:
According to question, we have
3^x + 3^(x+1) + 3^(x+2) = 13
Using the above formula, we have
3^x + 3^x*3 + 3^x*3^2 = 13
3^x(1 + 3 + 9) = 13
3^x(13) = 13
3^x = 13/13
3^x = 1
3^x = 3^0
So, the value of x is 0.
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Answers & Comments
Appropriate Question :- Solve
[tex]\sf \: 3^{x} + 3^{x + 1} + 3^{x + 2} = 13 \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given equation is
[tex]\sf \: 3^{x} + 3^{x + 1} + 3^{x + 2} = 13 \\ \\ [/tex]
We know,
[tex]\boxed{ \sf{ \: {x}^{m + n} = {x}^{m} \times {x}^{n} \: }} \\ \\ [/tex]
So, using this law of exponents, above equation can be rewritten as
[tex]\sf \: 3^{x} + 3^{x} \times {3}^{ 1} + 3^{x} \times {3}^{2} = 13 \\ \\ [/tex]
[tex]\sf \: 3^{x} (1+{3}^{ - 1}+{3}^{2}) = 13 \\ \\ [/tex]
[tex]\sf \: 3^{x} (1+3 + 9) = 13 \\ \\ [/tex]
[tex]\sf \: 3^{x} (1+3 + 9) = 13 \\ \\ [/tex]
[tex]\sf \: 3^{x} (13) = 13 \\ \\ [/tex]
[tex]\sf \: {3}^{x} = \dfrac{13}{13} \\ \\ [/tex]
[tex]\sf \: {3}^{x} = 1 \\ \\ [/tex]
[tex]\sf \: {3}^{x} = {3}^{0} \\ \\ [/tex]
[tex]\bf\implies \:x = 0 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
x = 0
Step-by-step explanation:
To Find:
Formula used:
According to question, we have
3^x + 3^(x+1) + 3^(x+2) = 13
Using the above formula, we have
3^x + 3^x*3 + 3^x*3^2 = 13
3^x(1 + 3 + 9) = 13
3^x(13) = 13
3^x = 13/13
3^x = 1
3^x = 3^0
x = 0
So, the value of x is 0.