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

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...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Replies
10
Views
2K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
1
Views
3K
Back
Top