Hello,
you can let a = 1.05 and b = 0.92 to make it simpler, then you have to solve :
300 a^x = 1000 b^x
Taking the logarithms of base "a" yields :
log_a(300 a^x) = log_a(1000 b^x)
Simplifying using the log laws :
log_a(300) + log_a(a^x) = log_a(1000) + log_a(b^x)
Keep simplifying :
log_a(300) + x = log_a(1000) + x log_a(b)
Isolate the x's on one side :
log_a(300) - log_a(1000) = x log_a(b) - x
Simplify :
log_a(300 / 1000) = x(log_a(b) - 1)
Alright :
(log_a(0.3)) / (log_a(b) - 1) = x
Now replace a and b with the values defined at the start, do the calculation and you got x (provided it exists, you'll know it doesn't if you get a math error )
EDIT : sorry, used 100 instead of 1000, corrected and checked against Wolfram