3(4^x)=250^x and 5^(x-2)=4^(2x+3) HELP PLEASE?

3 (4^x) = 250^x
log[3 (4^x)] = log(250^x)
log(3) + log(4^4) = x log(250)
log(3) + x log(4) = x log(250)
x log(250) - x log(4) = log(3)
x log(250/4) = log(3)
x = log(3)/log(62.5)


5^(x - 2) = 4^(2x + 3)
log[5^(x - 2)] = log[4^(2x + 3)]
(x - 2) log(5) = (2x + 3) log(4)
x log(5) - 2 log(5) = 2x log(4) + 3 log(4)
x log(5) - x log(4^2) = log(4^3) + log(5^2)
x log(5/16) = log(64*25)
x = log(1600)/log(0.3125)

3(4^x) = 250^x
(250^x) / (4^x) = 3
(250/4)^x = 3
62.5^x = 3
x = (ln(3) + 2*i*pi*n) / ln(62.5), for any integer n
x =~ 0.26567546278802995011819623353409 + 1.5194515676605343417262867213875*i*n, for any integer n
If x is a real number, then n = 0, so x = log[62.5](3) =~ 0.26567546278802995011819623353409
http://www.wolframalpha.com/input/?i=3%284^x%29%3D250^x

5^(x - 2) = 4^(2x + 3)
(5^x) / (5^2) = (4^(2x)) * (4^3)
(5^x) / 25 = (4^2)^x * 64
(5^x) = (16)^x * 64 * 25
(5^x) / (16^x) = 1600
(5/16)^x = 1600
0.3125^x = 1600
x = (ln(1600) + 2*i*pi*n) / ln(0.3125), for any integer n
x =~ -6.3429082850059839353806868350208 - 5.4018664254115702318363147656755*i*n, for any integer n
If x is a real number, then n = 0, so x = log[0.3125](1600) =~ -6.3429082850059839353806868350208
http://www.wolframalpha.com/input/?i=5^%28x+-+2%29+%3D+4^%282x+%2B+3%29

3 (4^x ) = 250^x
log(3) + x log(4) = x log(250)
x ( log(4) - log(250) = -log(3)
x = -log(3) / ( log(4) - log(250) )
x = 0.26568

5^(x-2) = 4^(2x+3)
(x-2) log(5) = (2x+3) log(4)
x log(5) - 2 log(5) = 2x log(4) + 3 log(4)
x log(5) - log(5^2) = 2x log(4) + log(4^3)
x log(5) -- log(25) = 2x log(4) + log(64)
x log(5) - 2x log(4) = log(25) + log(64)
x ( log(5) - 2 log(4) ) = log(25) + log(64)
x ( log(5) - log(4^2) ) = log(25) + log(64)
x ( log(5) - log(16) ) = log(25) + log(64)
x = ( log(25) + log(64) ) / ( log(5) - log(16) )
x = -6.3429