Question 24

If $$36^3 \times (4096 )^{\frac{1}{2}} \times 144 \times 9 \div (9^3 \times 72^2) = 4^x$$ , find the value of '$$x$$'.

$$36^3 \times (4096 )^{\frac{1}{2}} \times 144 \times 9 \div (9^3 \times 72^2) = 4^x$$

we can write this

$$36\times 36\times 36\times 64\times 144 \times 9 \div (9\times 9\times 9\times 72\times 72) = 4^x$$

after cancellation the numerator and denominator we get

$$64\times 16 =4^x$$

$$4^3\times 4^2 =4^x$$

$$x = 5$$

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