# Solve the equation 3^(2x + 1) = 70

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From the link provided the equation to be solved is `3^(2x + 1) = 70`

`3^(2x + 1) = 70`

=> `3^(2x)*3 = 70`

=> `9^x*3 = 70`

=> `9^x = 70/3`

take the logarithm of both the sides

`x*log 9 = log(70/3)`

=> `x = log(70/3)/(log 9)`

=> `x ~~ 1.4335`

**The approximate solution of the equation `3^(2x + 1) = 70` is `x = 1.4445` **

You should solve for the equation `3^(2x+1) = 70` , hence, using the properties of exponentials, you may write `3^(2x+1) = 3^(2x)*3` such that:

`3^(2x)*3 = 70 =gt 3^(2x) = 70/3`

You should take logarithms both sides such that:

`ln 3^(2x) = ln(70/3)`

`2x ln 3 = ln(70/3)`

`2x = ln(70/3)/(ln 3) =gt x = ln(70/3)/(2ln 3)=gt x=ln(70/3)/(ln 9)`

`x = (ln 70 - ln 3)/(ln 9)`

**Hence, evaluating the solution to exponential equation yields `x = (ln 70 - ln 3)/(ln 9).` **

**Sources:**

we have the euation as

3^(2x + 1) = 70

now taking log on both side

log (3^(2x + 1)) =log ( 70)

now log a^b = b log a

so

(2x+1) log 3 = log 70

2x+1 = log 70/log 3 = 1.845/0.477 = 3.868

2x+1=3.868

2x=3.868-1 = 2.868

x=2.868/2=1.434