Question

One ordered pair $(a,b)$ satisfies the two equations $ab^4 = 384$ and $a^2 b^5 = 4608$. What is the value of $a$ in this ordered pair?

Answer 1

Answer:

Therefore the value of a is 3.78.

Step-by-step explanation:

Indices Rule:

• $(a^m)^n=a^{mn}$
• $a^m.a^n=a^{m+n}$
• $\frac{a^m}{a^n}=a^{m-n}$
• $\sqrt[n]{a} =a^{\frac 1n}$
• $(\sqrt[n]{a} )^m=a^\frac mn$

Given that,

(a,b) satisfies the two equations

$ab^4=384$ .........(1)

 and    

$a^2b^5=4608$.........(2)

From equation (1) we get

$ab^4=384$

Squaring both sides

$\Rightarrow (ab^4)^2=(384)^2$

$\Rightarrow a^2b^8=147,456$......(3)

Divide equation (3) by (2) we get

$\frac{a^2b^8}{a^2b^5}=\frac{147,456}{4608}$

$\Rightarrow a^{2-2}b^{8-5}=32$

$\Rightarrow b^3=32$

$\Rightarrow b=\sqrt[3] {32}$

Now plug the value of b in equation (1)

$ab^4=384$

$\Rightarrow a(\sqrt[3]{32})^4=384$

$\Rightarrow a(32)^\frac43=384$

$\Rightarrow a=\frac{384}{(32)^\frac43}$

$\Rightarrow a\approx 3.78$

Therefore the value of a is 3.78.