3^(x+1) = 30, find x.?

3^x = 10 --->x = 1/log3 = 2.10

3^(x + 1) = 30
log[3^(x + 1)] = log(30)
(x + 1)log(3) = log(30)
(x)log(3) + log(3) = log(30)
(x)log(3) = log(30) - log(3)
x = [log(10)]/[log(3)]
x = 1/[log(3)]

3^(x+1) = 30 gives
3^x .3^1 = 30
or 3^x=10
Now take log of both the sides
x log3=log10
or x log3=1
or x/log3

( x + 1 ) log 3 = log 30
x + 1 = log 30 / log 3
x = ( log 30 / log 3 ) - 1
x = 2.096 (using any log base)

Note that 3^(x + 1) = (3)(3^x) So,

3^(x+1) = 30

<=>

3^x = 10

<=> take log (base 10) of both sides:

log(3^x) = log(10)

<=>

x log(3) = 1

<=>

x = 1/log(3) *** EXACT SOLUTION

≈ 2.096 *** APPROXIMATE SOLUTION

3^(x+1) = 30
(x+1)log3 = log30
x+1 = log30/log3
x = log30/log3 - 1

30^(1/3)-1=x=2.107232506