Hi, assuming the question as,log29-2x3-x=1log29-2x=3-x 9-2x=23-x=82xlet 2x= t 9-t = 8tt2-9t+8 = 0 t2-8t-t+8= 0 tt-8-1t-8= 0 t-8t-1= 0 t = 1, 8 2x= 1 or 2x= 82x= 20 or 2x= 23comparing we get x = 0 , 3 Regards

log2 (9-2^x)/ 3-x=1

Hi,

assuming the question as, \frac{\log_{2} (9 - 2^{x})}{3 - x} = 1 \log_{2} (9 - 2^{x}) = 3 - x 9 - 2^{x} = 2^{3 - x} = \frac{8}{2^{x}} let 2^{x} = t 9 - t = \frac{8}{t} t^{2} - 9 t + 8 = 0 t^{2} - 8 t - t + 8 = 0 t (t - 8) - 1 (t - 8) = 0 (t - 8) (t - 1) = 0 t = 1, 8 2^{x} = 1 or 2^{x} = 8 2^{x} = 2^{0} or 2^{x} = 2^{3} comparing we get x = 0, 3

Regards

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