Writing indices in a specific form.

[megz123]

Write \(\displaystyle \left(\frac{2^r\, \times\, 8^4}{2^{2r}\, \times\, 16}\right)\) in the form \(\displaystyle 2^{ar+b}\).

 

[JeffM]

Write \(\displaystyle \left(\frac{2^r\, \times\, 8^4}{2^{2r}\, \times\, 16}\right)\) in the form \(\displaystyle 2^{ar+b}\).

Please read READ BEFORE POSTING before posting again.

Generally, we provide help, not answers. Show us your work until where you are stuck, or try to explain what is stopping you from even starting.

Here is a hint to get you started: the KEY to this problem is to recognize that for the problem to be doable at all, 8 and 16 must be powers of 2. Do you see why?

Try restating the expression in terms of just powers of 2. What do you get?

What's the next step?

 

[megz123]

My apologies for not reading the READ BEFORE POSTING before I posted. I forgot to attach my work which I am stuck on because that was my first post. Thanks for the tips, but I still cannot work it out, help please?

\(\displaystyle \left(\dfrac{2^r\, \times\, 8^4}{2^{2r}\, \times\, 16}\right)\, =\, \dfrac{2^r\, \times\, 2^{3+r}}{2^{2r}\, \times\, 2^4}\, =\, \dfrac{2^{3+2r}}{2^{4+2r}}\, =\, \)

The answer provided was 2^2r-4.

Many thanks in advance.

 

[JeffM]

My apologies for not reading the READ BEFORE POSTING before I posted. I forgot to attach my work which I am stuck on because that was my first post. Thanks for the tips, but I still cannot work it out, help please?

\(\displaystyle \left(\dfrac{2^r\, \times\, 8^4}{2^{2r}\, \times\, 16}\right)\, =\,\) \(\displaystyle \dfrac{2^r\, \times\, 2^{3+r}}{2^{2r}\, \times\, 2^4}\) \(\displaystyle \, =\, \dfrac{2^{3+2r}}{2^{4+2r}}\, =\, \)

Your error is in the numerator of the fraction in red.

The answer provided was 2^2r-4.

Many thanks in advance.

Laws of exponents, which is what we call indices in the US. You need to memorize them.

\(\displaystyle a \ne 0 \implies a^0 \equiv 1.\)

\(\displaystyle a^1 \equiv a.\)

\(\displaystyle a^b * a^c \equiv a^{(b + c)}.\)

\(\displaystyle a^b \div a^c \equiv a^{(b - c)}.\)

\(\displaystyle a \ne 0\ \implies \ a^{(-b)} \equiv \dfrac{1}{a^b} \equiv \left(\dfrac{1}{a}\right)^b.\)

\(\displaystyle \left(a^b\right)^c \equiv a^{(bc)} \equiv \left(a^c\right)^b.\)

\(\displaystyle a^{(b/c)} \equiv \sqrt[c]{a^b}.\)

You actually understood my hint, but then misapplied the laws of exponents.

\(\displaystyle 8 = 2^3 \implies 8^4 = \left(2^3\right)^4 = what?\) There is no r in the answer. Look for the applicable law.

\(\displaystyle 16 = 2^4.\) See you got the hint.

Do you see your error now? Can you complete it on your own now?

 

[Deleted member 4993]

My apologies for not reading the READ BEFORE POSTING before I posted. I forgot to attach my work which I am stuck on because that was my first post. Thanks for the tips, but I still cannot work it out, help please?

\(\displaystyle \left(\dfrac{2^r\, \times\, 8^4}{2^{2r}\, \times\, 16}\right)\, =\, \dfrac{2^r\, \times\, 2^{3+r}}{2^{2r}\, \times\, 2^4}\, =\, \dfrac{2^{3+2r}}{2^{4+2r}}\, =\, \)

The answer provided was 2^2r-4.

Many thanks in advance.

Looking at the answer - I believe the problem is:

\(\displaystyle \left(\dfrac{2^r\, \times\, 8^r}{2^{2r}\, \times\, 16}\right)\, =\, \dfrac{2^r\, \times\, 2^{3*r}}{2^{2r}\, \times\, 2^4}\, =\, \dfrac{2^{4r}}{2^{4+2r}}\, =\, ??\)

 

[megz123]

Ohh so 8^r = (2³)^r ?

I think I got it.

Thanks heaps!