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

2 Answers
Jun 30, 2017

Answer:

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#

Jun 30, 2017

Answer:

#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"#