How would you solve this? i think with logs

but how.....?

2^(x-1) = 10

2^(x - 1) = 10
log(base 2) 2^(x - 1) = log(base 2) 10
x - 1 = log(base 2) 10

change the base of the logarithm:
log(base a) x = log(base b) x / log(base b) a
log(base 2) 10 = log(base 10) 10 / log(base 10) 2
when the base is 10, we can leave off the 10 in the notation

x - 1 = log(base 2) 10
x - 1 = (log 10)/(log 2)
x - 1 = 3.32
x = 4.32

2^(x-1) = 10

take the Log₂ of both sides

=> x-1 = Log₂(10)
=> x = Log₂(10) + 1

given Log(ab) = Log (a) + Log (b)
=> x = Log₂(2) + Log₂(5) + 1

given Log [base a] (a^x) = x
=> x =1 + Log₂(5) + 1
=> x = Log₂(5) + 2

if you don't have a Log₂ look up table, or can't do Log₂ on you calculator

=> x = Log(5)/Log(2) + 2 in any base will give you the same answer (most scientific calculators have Log base 10 or Log base e (ln) functions)

=> x = 4.321928 (6 dp)

(x-1)log2 = log10
(x-1)(.3010) = 1
x-1 = 3.322
x = 4.322

Hint I think the answer is reapeting