Question

solve:log16 x+log4 x+log2 x=7​

Answer 1

Step-by-step explanation:

logx=7(1-log2)/3 is simplified

Answer 2

Answer:

Log16x + log4x + log2x = 7

use change of base formula

logx/log16 + logx/log4 + logx/log2 = 7

logx/log(2^4) + logx/log(2^2) + logx/log2 = 7

logx/4log2 + logx/2log2 + logx/log2 = 7

logx/log2 * (1/4 + 1/2 + 1) = 7

logx/log2 * (7/4) = 7

logx/log2 * (1/4) = 1

logx/log2 = 4

logx = 4log2

logx = log(2^4)

x = 2^4 = 16

Step-by-step explanation:

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