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    Solve for x

    You're right about going around in circles. (x) 1= \frac{2^{x+1}}{x} (x) x= 2^{x+1} \ln(x)= \ln(2^{x+1}) \ln(x)= x \ln(2) + \ln(2) \ln(x) - \ln(2) = x \ln(2) \frac{\ln(x)}{\ln(2)} - 1 = x This will give you an x = something, but you still have x in both sides of the...
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