Appropriate Question :-
[tex]\sf \: If \: {0.357}^{a} = {357}^{b} = {1000}^{c}, \: prove \: that \: b(a + c) = ac \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \: {0.357}^{a} = {357}^{b} = {1000}^{c} \: \\ \\ [/tex]
Let assume that
[tex]\sf \: \: {0.357}^{a} = {357}^{b} = {1000}^{c} = k\: \\ \\ [/tex]
So,
[tex]\sf \: \: {0.357}^{a} = k\:\sf \: \implies \: 0.357 = {\bigg(k\bigg) }^{ \dfrac{1}{a} } \\ \\ [/tex]
Also,
[tex]\sf \: \: {357}^{b} = k\:\sf \: \implies \: 357 = {\bigg(k\bigg) }^{ \dfrac{1}{b} } \\ \\ [/tex]
[tex]\sf \: \: {1000}^{c} = k\:\sf \: \implies \: 1000 = {\bigg(k\bigg) }^{ \dfrac{1}{c} } \\ \\ [/tex]
Now,
[tex]\sf \: 0.357 \times 1000 = 357 \\ \\ [/tex]
On substituting the values from above, we get
[tex]\sf \: {\bigg(k\bigg) }^{ \dfrac{1}{a} } \times {\bigg(k\bigg) }^{ \dfrac{1}{c} } = {\bigg(k\bigg) }^{ \dfrac{1}{b} } \\ \\ [/tex]
[tex]\sf \: {\bigg(k\bigg) }^{ \dfrac{1}{c} + \dfrac{1}{a} } = {\bigg(k\bigg) }^{ \dfrac{1}{b} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {x}^{m} \times {x}^{n} \: = \: {x}^{m + n} \: }} \\ \\ [/tex]
[tex]\sf \: {\bigg(k\bigg) }^{ \dfrac{a + c}{ac} } = {\bigg(k\bigg) }^{ \dfrac{1}{b} } \\ \\ [/tex]
[tex]\sf \: \dfrac{a + c}{ac} = \dfrac{1}{b} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {x}^{m} = {x}^{n} \:\sf \: \implies \: m = n\: }} \\ \\ [/tex]
[tex]\bf\implies \:b(a + c) = ac \\ \\ \\ [/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]
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Verified answer
Appropriate Question :-
[tex]\sf \: If \: {0.357}^{a} = {357}^{b} = {1000}^{c}, \: prove \: that \: b(a + c) = ac \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \: {0.357}^{a} = {357}^{b} = {1000}^{c} \: \\ \\ [/tex]
Let assume that
[tex]\sf \: \: {0.357}^{a} = {357}^{b} = {1000}^{c} = k\: \\ \\ [/tex]
So,
[tex]\sf \: \: {0.357}^{a} = k\:\sf \: \implies \: 0.357 = {\bigg(k\bigg) }^{ \dfrac{1}{a} } \\ \\ [/tex]
Also,
[tex]\sf \: \: {357}^{b} = k\:\sf \: \implies \: 357 = {\bigg(k\bigg) }^{ \dfrac{1}{b} } \\ \\ [/tex]
Also,
[tex]\sf \: \: {1000}^{c} = k\:\sf \: \implies \: 1000 = {\bigg(k\bigg) }^{ \dfrac{1}{c} } \\ \\ [/tex]
Now,
[tex]\sf \: 0.357 \times 1000 = 357 \\ \\ [/tex]
On substituting the values from above, we get
[tex]\sf \: {\bigg(k\bigg) }^{ \dfrac{1}{a} } \times {\bigg(k\bigg) }^{ \dfrac{1}{c} } = {\bigg(k\bigg) }^{ \dfrac{1}{b} } \\ \\ [/tex]
[tex]\sf \: {\bigg(k\bigg) }^{ \dfrac{1}{c} + \dfrac{1}{a} } = {\bigg(k\bigg) }^{ \dfrac{1}{b} } \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {x}^{m} \times {x}^{n} \: = \: {x}^{m + n} \: }} \\ \\ [/tex]
[tex]\sf \: {\bigg(k\bigg) }^{ \dfrac{a + c}{ac} } = {\bigg(k\bigg) }^{ \dfrac{1}{b} } \\ \\ [/tex]
[tex]\sf \: \dfrac{a + c}{ac} = \dfrac{1}{b} \\ \\ [/tex]
[tex]\boxed{ \sf{ \: \because \: {x}^{m} = {x}^{n} \:\sf \: \implies \: m = n\: }} \\ \\ [/tex]
[tex]\bf\implies \:b(a + c) = ac \\ \\ \\ [/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]