MHB Where Did I Go Wrong When Simplifying This Exponent Equation?

Click For Summary
The discussion centers around the simplification of the expression \[ \left[(-4a^{-4}b^{-5})^{-3}\right]^4 \]. The initial mistake was in incorrectly applying the exponent of -3 to the constant -4, leading to an erroneous calculation of the final result. The correct approach involves recognizing that the negative sign must be accounted for, resulting in a negative fraction when simplified. Ultimately, the correct answer is \[ \frac{a^{48}b^{60}}{16777216} \], as confirmed by an online calculator. Understanding the treatment of negative exponents and signs is crucial for accurate simplification.
PistolSlap
Messages
2
Reaction score
0
I have this problem to simplify with positive exponents:

\[\left[(-4a^{-4}b^{-5})^{-3}\right]^4\]

So, working with the interior brackets, I applied -3 to the equation, which resulted in:

\[-(-64)x^{12}b^{15}\]

**because "-4" was not in brackets, the exponent was applied to the 4 only, independent of the negative sign, which resulted in -(-64), so:

\[\left[ 64x^{12}b^{15}\right]^4\]

which resulted in this insane answer:

\[16777216x^{48}b^{60}\]

However, when I checked it, an online calculator said the answer was:

\[\frac{a^{48}b^{60}}{16777216}\]

which means that when applying exponents to the number I did something wrong, as it should have ended up negative, which would have resulted in it becoming a positive denominator.

What did I do wrong to end up with the wrong sign on that number?
 
Last edited by a moderator:
Mathematics news on Phys.org
Re: Simplifying Question -- Please Help?

In your first step, you incorrectly applied the exponent of -3 to the constant -4, you applied a positive 3 instead. Your first step should look like:

$\displaystyle \left[\left(-4a^{-4}b^{-5} \right)^{-3} \right]^4=\left[(-4)^{1(-3)}a^{-4(-3)}b^{-5(-3)} \right]^4=\left[(-4)^{-3}a^{12}b^{15} \right]^4=\left[-\frac{a^{12}b^{15}}{4^3} \right]^4$

Now, you can see why the result is as given by the online calculator you used.
 
the various negative signs make this kind of complicated.

the first thing i would do is recognize that:

$-4a^{-4}b^{-5} = (-1)(4)(a^{-4}b^{-5})$

so:

$(-4a^{-4}b^{-5})^{-3} = (-1)^{-3}(4)^{-3}(a^{-4})^{-3}(b^{-5})^{-3}$

and:

$(-1)^{-3} = \dfrac{1}{(-1)^3} = \dfrac{1}{-1} = -1$, so

$(-1)^{-3}(4)^{-3}(a^{-4})^{-3}(b^{-5})^{-3} = -[(4)^{-3}(a^{-4})^{-3}(b^{-5})^{-3}]$

now, "inside the brackets" the first term is:

$4^{-3} = \dfrac{1}{4^3}$, so we have:

$(-4a^{-3}b^{-5})^{-3} = -\left(\dfrac{a^{(-4)(-3)}b^{(-5)(-3)}}{4^3}\right) = -\left(\dfrac{a^{12}b^{15}}{4^3}\right)$

taking the 4th power of this, the negative sign goes away, and we get:

$[(-4a^{-3}b^{-5})^{-3}]^4 = \left[-\left(\dfrac{a^{12}b^{15}}{4^3}\right)\right]^4 = \left(\dfrac{a^{12}b^{15}}{4^3}\right)^4$

$ = \dfrac{a^{48}b^{60}}{4^{12}}$

as a side note, your reasoning that since (-4) was "not in brackets" the exponent did not apply to the negative sign but only to the 4 is wrong...you just got lucky, because -3 is ODD.

for example:

$(-2a)^2 = 4a^2$ but $-(2a)^2 = -4a^2$

since the first is $(-2a)(-2a)$ while the second is $-(2a)(2a)$.
 
Re: Simplifying Question -- Please Help?

Awesome, thanks, I understand now! :D

MarkFL said:
In your first step, you incorrectly applied the exponent of -3 to the constant -4, you applied a positive 3 instead. Your first step should look like:

$\displaystyle \left[\left(-4a^{-4}b^{-5} \right)^{-3} \right]^4=\left[(-4)^{1(-3)}a^{-4(-3)}b^{-5(-3)} \right]^4=\left[(-4)^{-3}a^{12}b^{15} \right]^4=\left[-\frac{a^{12}b^{15}}{4^3} \right]^4$

Now, you can see why the result is as given by the online calculator you used.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K