Solve for $d-b$: $a^5=b^4,\,c^3=d^2,\,c-a=19$

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In summary, the value of $d-b$ can be determined by solving for $d$ and $b$ in the given equations and subtracting $b$ from $d$. The equations can be solved for $d-b$ by first solving for $d$ and $b$ and then subtracting $b$ from $d$, resulting in a numerical value for $d-b$. There is no specific method for solving for $d-b$ in these equations, but one approach could involve solving for $c$ in terms of $a$ and then substituting that into the equation $c^3=d^2$ to find the value of $d$. The equations can be solved without using numerical values by algebraically manipulating them
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anemone
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Assume that $a,\,b,\,c$ and $d$ are positive integers such that $a^5=b^4,\,c^3=d^2$ and $c-a=19$.

Determine $d-b$.
 
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anemone said:
Assume that $a,\,b,\,c$ and $d$ are positive integers such that $a^5=b^4,\,c^3=d^2$ and $c-a=19$.

Determine $d-b$.

As $c^3=d^2$ So there exists x such that $c=x^2$ and $d = x^3$
Further as $a^5=b^4$ so there exists y such that $a=y^4$ and $b=y^5$
Now
c-a=19
$=> x^2-y^4=19$
$=> (x-y^2)(x+y^2)=19$
As 19 is prime and $x+y^2 > x- y^2$ we have
$x-y^2=1$ and $x+y^2=19$
Solving these we get $x=10$ and $y = 3$
So $d-b = 10^3 - 3^5 = 1000 - 243= 757$
 

FAQ: Solve for $d-b$: $a^5=b^4,\,c^3=d^2,\,c-a=19$

How do I solve for $d-b$ in this equation?

To solve for $d-b$, we need to isolate the variable on one side of the equation. In this case, we can use the given equations to substitute values for $a$, $b$, and $c$ in terms of $d$. Then, we can solve for $d$ and plug the value back into the equation to find $b$. Finally, we can subtract $b$ from $d$ to get the value of $d-b$.

Can you explain the steps to solving this equation?

First, we need to rearrange the equations to isolate $d$ on one side. We can use the fact that $a^5=b^4$ to substitute $a$ in terms of $b$ in the second equation. Then, we can substitute $d$ in terms of $c$ in the first equation. Next, we can use the third equation to substitute $a$ in terms of $c$. Finally, we can solve for $d$ and plug the value back into the equation to find $b$. Subtracting $b$ from $d$ will give us the value of $d-b$.

Is there a simpler way to solve this equation?

There may be multiple methods to solve this equation, but the given steps are a systematic approach to finding the value of $d-b$. It may be possible to simplify the equations or use different substitutions, but the end result will still be the same.

Can you provide an example of solving for $d-b$ using this equation?

Sure, let's say $a=2$, $b=4$, $c=27$. We can substitute these values into the equations to get $2^5=4^4$, $27^3=d^2$, and $27-2=19$. From the first equation, we know that $a^5=b^4$ and $2^5=32$, so $b^4=32$ and $b=2$. Substituting this into the second equation, we get $27^3=d^2$ and $d=27$. Finally, we can subtract $2$ from $27$ to get the value of $d-b$, which is $25$.

Are there any restrictions or limitations to solving this equation?

There may be certain restrictions or limitations depending on the given values in the equations. For example, if $a$ and $b$ are both negative, then the equation $a^5=b^4$ may not hold true. Additionally, if the given values do not result in a real solution for $d$, then the equation may not have a solution for $d-b$.

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