Question

How do you multiply and simplify `\frac { 6b } { 4b ^ { 2} } \cdot \frac { 8b ^ { 4} } { 9b ^ { 5} }`?

Answer 1

See a solution process below:

Explanation:

First, rewrite this expression as:

`(6 * 8)/(4 * 9) * (b * b^4)/(b^2 * b^5)`

Next, factor and cancel common terms in the contants:

`(3 * 2 * 4 * 2)/(4 * 3 * 3) * (b * b^4)/(b^2 * b^5) =>`

`(color(red)(cancel(color(black)(3))) * 2 * color(blue)(cancel(color(black)(4))) * 2)/(color(blue)(cancel(color(black)(4))) * color(red)(cancel(color(black)(3))) * 3) * (b * b^4)/(b^2 * b^5) =>`

`(2 * 2)/3 * (b * b^4)/(b^2 * b^5) =>`

`4/3 * (b * b^4)/(b^2 * b^5)`

Now, use these rules of exponents to simplify the `b` terms in the numerator:

`a = a^color(red)(1)` and `x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))`

`4/3 * (b^color(red)(1) xx b^color(blue)(4) * b^4)/(b^2 * b^5) => 4/3 * (b^(color(red)(1) + color(blue)(4)))/(b^2 * b^5) => 4/3 * b^5/(b^2 * b^5) =>`

`4/3 * color(red)(cancel(color(black)(b^5)))/(b^2 * color(red)(cancel(color(black)(b^5)))) =>`

`4/(3b^2)`

Answer 2

`(6b)/(4b^2)*(8b^4)/(9b^5)` simplifies to `color(blue)(4/(3b^2)`.

Explanation:

Multiply and simplify.

`(6b)/(4b^2)*(8b^4)/(9b^5)`

Multiply the numerators and denominators across the two fractions.

`(6bxx8b^4)/(4b^2xx9b^5)`

Multiply the coefficients.

`(6xx8xxbxxb^4)/(4xx9xxb^2xxb^5)`

Simplify.

`(48xxbxxb^4)/(36xxb^2xxb^5)`

Apply exponent product rule: `x^mx^n=x^(m+n)`.

`(48xxb^(1+4))/(36xxb^(2+5))`

Simplify.

`(48b^5)/(36b^7)`

Simplify `48/36`.

`((48-:12)b^5)/((36-:12)b^7)`

`(4b^5)/(3b^7)`

Apply quotient exponent rule: `x^m/x^n=x^(m-n)`.

`(4b^(5-7))/3`

Simplify.

`(4b^(-2))/3`

Apply the negative exponent rule: `x^(-m)=1/x^m`.

`4/(3b^2)`

Answer 3

`4/(3b^2)`

Explanation:

Let's start with the original problem:

`(6b)/(4b^2)*(8b^4)/(9b^5)`

Let's simplify the expression by rewriting the expression and canceling out some terms:

`6/4*b/b^2*8/9*b^4/b^5=>cancel6^3/cancel4^2*b/b^2*8/9*b^4/b^5`

`8/9` can't be canceled so we will leave it as is. However, we can simplify the variable exponents by using one of the rules pertaining to exponents:

`a^-n=1/a^n`

Using this, we can simplify the expression:

`3/2*b/b^2*8/9*b^4/b^5=>3/2*b^-1*8/9*b^-1=>3/2*1/b*8/9*1/b`

We can then rearrange the expression like so:

`3/2*1/b*8/9*1/b=>3/2*8/9*1/b*1/b`

We notice that we can cancel out the `3` and the `9` and the `2` and the `8` in the two terms at the front:

`3/2*8/9*1/b*1/b=>cancel3^1/cancel2^1*cancel8^4/cancel9^3*1/b*1/b`

Then, we multiply the numerical terms and the variable terms together to obtain our answer:

`1/1*4/3*1/b*1/b=>4/3*1/b^2=>4/(3b^2)`