Question

How do you solve `5^21 times 4^11 = 2 times 10^n`?

Answer 1

`n=21`

Explanation:

As `5^21xx4^11=2xx10^n` hence

`21log5+11log4=log2+n`

Hence `n=21log5+11log2^2-log2`

= `21log5+22log2-log2`

= `21log5+21log2`

= `21(log5+log2)`

= `21log10=21`

Answer 2

`n = 21`

Explanation:

`5^21 * 4^11 = 2 * 10^n`

`(10/2)^21 * (2^2)^11 = 2 * 10^n`

`10^{21} * color{red}{(1/2)^21 * (2)^22} = 2 * 10^n`

`10^{21} * 2 = 2 * 10^n`

`n = 21`

Answer 3

`n=21`

Explanation:

`5^{21}cdot 4^{11} = 2 cdot 10^n`
`5^{10}cdot 5^{10}cdot 4^{10} cdot 5 cdot 4 = 2 cdot 10^n`
`(5 cdot 5 cdot 4)^10cdot 5 cdot 4 = 2 cdot 10^n`
`100^10cdot 20 = 2 cdot 10^n`
`10^{20}cdot 20 = 2 cdot 10^n`
`2cdot 10^{21} = 2 cdot 10^n`
`n = 21`