Question

if 8 ki power x+1 =64 what is the value of 3 ki Power 2x+1
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Answer 1

CORRECT QUESTION :-

$\mapsto \sf \: if \: 8 ^{x + 1} = 64 \: then \: find \: the \: value \: of \: 3 ^{2x + 1}$

GIVEN :-

$\mapsto \sf \: 8 ^{x + 1} = 64$

TO FIND :-

$\mapsto \sf \: the \: value \: of \: \: 3 ^{2x + 1}$

SOLUTION :-

$\mapsto \sf \: 8 ^{x + 1} = 64$

• Now expand 64 as 8².

$\mapsto \sf \:( 8) ^{x + 1} = (8) ^{2}$

• Now on comparing bases , The exponents will be also equal.

$\mapsto \sf \: x + 1 = 2 \\ \\ \mapsto \sf \: x = 2 - 1 \\ \\ \mapsto \underline{ \boxed{ \red{ \sf \: x = 1}}}$

Hence the value of x is 1.

• Now substitute the value of x in 3^2x + 1.

$\mapsto \sf \: (3) ^{2x + 1} \\ \\ \mapsto \sf \: (3) ^{2 \times 1 + 1} \\ \\ \mapsto \sf (3) ^{2 + 1} \\ \\ \mapsto \sf (3) ^{3} \\ \\ \mapsto \underline{ \boxed{ \red{ \sf \: (3) ^{2x + 1 } = 27}}}$

Hence the required answer is 27.

ADDITIONAL INFORMATION :-

$\boxed{\begin{minipage}{7cm} \\ \sf{ \implies \bf \sqrt[n]{ \sqrt[m]{ \sqrt[p]{((a^{x} )^{y}) ^{z} } } } = (a ^{xyz} )^{ \frac{1}{mnp} } = a ^{ \frac{xyz}{mnp} } } \\ \\ \sf{ \implies a^m \times a^n = a^{m+n} } \\ \\ \sf{ \implies {a}^{m} \times b^m = ab^m } \\ \\ \sf{ \implies \dfrac{a^m}{a^n} = a^{m - n} ( \tt{ If \: m > n} ) } \\ \\ \sf{ \implies \dfrac{a^m}{ a^n} = \dfrac{ 1}{ a^{n-m} } ( \tt{ If \: n > m )} } \\ \\ \sf{ \implies (a^m)^n = a^{mn} } \\ \\ \sf{ \implies a^{-n} = \dfrac{1}{ a^n} }\end{minipage}}$