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Question 26
If m and n are integers such that $$(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$$ then m is
Solution
We have, $$(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$$
Converting both sides in powers of 2 and 3, we get
$$2^{\ \frac{19\ }{2}}3^42^43^{2m}2^{3n}$$ = $$3^n2^{4m}2^{\frac{\ 6}{4}}$$
Comparing the power of 2 we get, $$\ \frac{\ 19}{2}+4+3n\ =4m+\frac{\ 6}{4}\ $$
=> 4m=3n+12 .....(1)
Comparing the power of 3 we get, $$4+2m=n$$
Substituting the value of n in (1), we get
4m=3(4+2m)+12
=> m=-12
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