Question

How do you solve `3^(2x+1) = 5`?

Answer 1

I got:
`x=1/2[ln(5)/ln(3)-1]=0.23486`

Explanation:

We take the natural log of both sides:

`ln(3)^(2x+1)=ln(5)`

apply a property of logs and write:

`(2x+1)ln(3)=ln(5)`

rearrange:

`2x+1=ln(5)/ln(3)`

`x=1/2[ln(5)/ln(3)-1]=0.23486`

Answer 2

`x~~0.232" to 3 dec. places"`

Explanation:

`"using the "color(blue)"law of logarithms"`

`• logx^nhArrnlogx`

`3^(2x+1)=5`

`"take ln (natural log) of both sides"`

`rArrln3^(2x+1)=ln5`

`rArr(2x+1)ln3=ln5`

`rArr2x+1=ln5/ln3larr" subtract 1 from both sides"`

`rArr2x=(ln5/ln3)-1larr" divide both sides by 2"`

`rArrx=1/2[(ln5/ln3)-1]~~0.232" 3 dec. places"`