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Question: If \[\sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } } = {\left( {343} \right)^{y - 1}}\] the...

If 7777.................=(343)y−1\sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } } = {\left( {343} \right)^{y - 1}} then yy is equal to
A. 23\dfrac{2}{3}
B. 11
C. 43\dfrac{4}{3}
D. 34\dfrac{3}{4}

Explanation

Solution

In this question, first of all consider the value of 7777.................\sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } } as a variable and then square the terms on both sides to find the value of the variable. Then substitute this value in the given data to get the required answer. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer :

Given that 7777.................=(343)y−1...................................(1)\sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } } = {\left( {343} \right)^{y - 1}}...................................\left( 1 \right)
Let x=7777.................x = \sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } }
On squaring both sides we have

⇒x2=(7777.................)2 ⇒x2=7(7777.................) ⇒x2=7x [∵x=7777.................]  \Rightarrow {x^2} = {\left( {\sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } } } \right)^2} \\\ \Rightarrow {x^2} = 7\left( {\sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } } } \right) \\\ \Rightarrow {x^2} = 7x{\text{ }}\left[ {\because x = \sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } } } \right] \\\

Cancelling the common terms, we have
∴x=7\therefore x = 7
So, we have x=7777.................=7.......................................(2)x = \sqrt {7\sqrt {7\sqrt {7\sqrt {7.................} } } } = 7.......................................\left( 2 \right)
From equation (1) and (2), we get

⇒7=(343)y−1 ⇒7=(73)y−1 [∵343=73] ⇒7=(7)3(y−1) ⇒71=73y−3  \Rightarrow 7 = {\left( {343} \right)^{y - 1}} \\\ \Rightarrow 7 = {\left( {{7^3}} \right)^{y - 1}}{\text{ }}\left[ {\because 343 = {7^3}} \right] \\\ \Rightarrow 7 = {\left( 7 \right)^{3\left( {y - 1} \right)}} \\\ \Rightarrow {7^1} = {7^{3y - 3}} \\\

Comparing the powers as bases are equal, we have

⇒1=3y−3 ⇒1+3=3y ⇒4=3y ∴y=43  \Rightarrow 1 = 3y - 3 \\\ \Rightarrow 1 + 3 = 3y \\\ \Rightarrow 4 = 3y \\\ \therefore y = \dfrac{4}{3} \\\

Thus, the correct option is C. 43\dfrac{4}{3}

Note : Here, we can equate the powers of the terms in the equation 71=73y−3{7^1} = {7^{3y - 3}} as their bases are equal. To solve these kinds of questions, first consider the given surd as a variable and square them on both sides whenever we get a surd in terms or square roots and apply a cubing on both sides whenever we get a surd in terms of cubic roots.