log16 x 10g4 x log2 x 7 ex ex2 1 log5 x log3 x 1Solut

log_16 x + 10g_4 x + log_2 x = 7 e^x + e^-x/2 = 1 log_5 x + log_3 x = 1

Solution

log₁₆x+ log₄x+log₂x=7

using change of base rule

log₂x/log₂16 + log₂x/log₂4 +log₂x=7

log₂x/ 4 + log₂x/2 + log₂x=7

log₂x + 2log₂x + 4log₂x=28

7log₂x=28

log₂x=4

x=2⁴=16