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If 2^{a} = √64 and b/a=3, evaluate a^{2} + b^{2}

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Hello

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2^{a} = √64 ----------- eqn(1)

b/a=3 ------------ eqn(2),

From eqn(1),

2^{a} = 8

2^{a} = 2^{3}

From laws of indices,

a = 3

Substitute the value of a into eqn(2)

b/3 = 3

b = 3 × 3

b = 9

Therefore,

a^{2} + b^{2} = 3^{2} + 9^{2}

= 9 + 81

= 90

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2a = √64 ----------- eqn(1)

b/a=3 -------- eqn(2)

from eqn(2), multiply both sides by a,

b = 3a ---------- eqn(3)

Also, from eqn(1), 2a = 8

divide both sides by 2,

a = 8/2

**a = 4**

substitute a = 4 into eqn(3)

b = 3(4)

**b = 12**

Thus, a^{2} + b^{2} = 4^{2} + 12^{2}

= 16 + 144

= 160

Please how did you get a=4, remember the a is the square of the 2, not 2a

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