solve for x i need help with this question and i need the answer now
Is 4^x2 = 13 the same as 4^(x^2) = 13 or 4^(2x) = 13 ?
In either case, we can change each expression into a logarithm format to solve.
In general for converting to a log, we can say B^(P) = W
where B is the base, P is the power, and W is whatever. This can be converted to log as
logB(W) = P
where B is the base of the logarithm. For both cases, they can be written in log form as
log4(13) = x^(2) or log4(13) = 2x
Since 4 is a different than log base 10 which is the log function on your standard calculator, we can use the change of base formula which will convert to base 10. That is
loga(c) = log10(c)/log10(a) = log(c)/log(a) , where a is the base of the original logarithm.
For the two cases above, we'll have log4(13) = log(13)/log(4) = 1.11/0.60 [rounded to hundredths] = 1.85
So, we'll have 1.85 = x^(2) or 1.85 = 2x
For the first case, 1.85 = x^(2), x = +/- sqrt(1.85) where sqrt(x) means square root of x. That is
x = + 1.36, -1.36 [rounded to hundredths]
For the second case, 1.85 = 2x, we have x = 1.85/2 = 0.925
Hope this helps!