Question

How do you solve `4(.3)^x = 1.2^(x+2)`?

Answer 1

`(ln 5 - ln 3)/ (ln 2)`

Explanation:

take logs

`ln ( 4 (0.3)^x) = ln(1.2^{x+2})`
`ln ( 4 ) + ln (0.3^x) = ln(1.2^{x+2})`
`ln ( 4 ) + x ln (0.3) = (x+2) ln(1.2)`
`x (ln (0.3) - ln(1.2) )= 2 ln(1.2) - ln ( 4 )`
`x ln (0.3 / 1.2 )= ln(1.2^2 / 4)`
`x ln (1/4 )= ln(9/25)`
`x = ln(9/25) / ln (1/4 ) = (ln 9 - ln 25)/(ln 1 - ln 4)`
`= (2 ln 3 - 2 ln 5)/(0 - 2 ln 2)`
`= (ln 5 - ln 3)/ (ln 2)`

Answer 2

`x = 0.736966`

Explanation:

`4(.3)^x = 1.2^(x+2) = (4 xx .3)^{x+2} = 4^{x+2}xx(.3)^{x+2}`

then

`4(.3)^x = 4^{x+2}xx(.3)^{x+2} = 4^2xx(0.3)^2xx4^x xx(.3)^x`

elliminating `(.3)^x` in both sides

`4 = 4^2xx(0.3)^2xx4^x->4^x=1/(4 xx (0.3)^2)`

and finally

`x = -log_e(4 xx (0.3)^2)/log_e 4 = 0.736966`