Assume that a,b,c and d are positive integers such that a^5=b⁴,c³=d

1.	Assume that a,b,c and d are positive integers such that a^5=b⁴,c³=d²,and c-a=19 find d-b
Answer» $\large\underline{\sf{Solution-}}$ Given that, a, b, C, d are POSITIVE integers such that $\rm :\longmapsto\: {a}^{5} = {b}^{4}$ and $\rm :\longmapsto\: {c}^{3} = {d}^{2}$ As $\rm :\longmapsto\: {a}^{5} = {b}^{4}$ $\rm \implies\: {a}^{5} \: must \: be \: a \: perfect \: square$ Also, $\rm :\longmapsto\: {c}^{3} = {d}^{2}$ $\rm \implies\: {c}^{3} \: must \: be \: a \: perfect \: square$ Again, it is given that $\rm :\longmapsto\:c \: - \: a = 19$ Now, by hit and trial method, we have to choose values of c and a in such a way that they must be perfect SQUARES and difference is 19. So, By hit and trial $\bf :\longmapsto\:c \: = \: 100 \: and \: a \: = \: 81$ Now, $\rm :\longmapsto\: {a}^{5} = {b}^{4}$ On substituting the value of a, we get $\rm :\longmapsto\: {81}^{5} = {b}^{4}$ $\rm :\longmapsto\: {( {3}^{4} )}^{5} = {b}^{4}$ $\rm :\longmapsto\: {3}^{20} = {b}^{4}$ $\bf\implies \:b = {3}^{5} \: \: \: \: \: as \: b > 0$ $\bf\implies \:b = 243$ Also, $\rm :\longmapsto\: {c}^{3} = {d}^{2}$ On substituting the value of c, we get $\rm :\longmapsto\: {100}^{3} = {d}^{2}$ $\rm :\longmapsto\: {( {10}^{2} )}^{3} = {d}^{2}$ $\rm :\longmapsto\: {10}^{6} = {d}^{2}$ $\bf\implies \:d = {10}^{3} \: \: \: \: \: as \: d > 0$ $\bf\implies \:d = 1000$ Now, CONSIDER $\rm :\longmapsto\:d - b$ $\rm \: = \: 1000 - 243$ $\rm \: = \: 757$ Hence, $\red{\rm :\longmapsto\:\boxed{ \tt{ \: d \: - \: b \: = \: 757 \: }}}$ ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ Additional Information :- $\boxed{ \tt{ \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}$ $\boxed{ \tt{ \: {x}^{m} \div {x}^{n} = {x}^{m - n} \: }}$ $\boxed{ \tt{ \: {( {x}^{m} )}^{n} = {( {x}^{n} )}^{m} = {x}^{mn} \: }}$ $\boxed{ \tt{ \: {x}^{0} = 1 \: }}$ $\boxed{ \tt{ \: {x}^{ - n} = \frac{1}{ {x}^{n} } \: }}$

Assume that a,b,c and d are positive integers such that a^5=b⁴,c³=d²,and c-a=19 find d-b

Answer»

$\large\underline{\sf{Solution-}}$

Given that, a, b, C, d are POSITIVE integers such that

$\rm :\longmapsto\: {a}^{5} = {b}^{4}$

and

$\rm :\longmapsto\: {c}^{3} = {d}^{2}$

$\rm :\longmapsto\: {a}^{5} = {b}^{4}$

$\rm \implies\: {a}^{5} \: must \: be \: a \: perfect \: square$

Also,

$\rm :\longmapsto\: {c}^{3} = {d}^{2}$

$\rm \implies\: {c}^{3} \: must \: be \: a \: perfect \: square$

Again, it is given that

$\rm :\longmapsto\:c \: - \: a = 19$

Now, by hit and trial method, we have to choose values of c and a in such a way that they must be perfect SQUARES and difference is 19.

So, By hit and trial

$\bf :\longmapsto\:c \: = \: 100 \: and \: a \: = \: 81$