Question

logx4+logx16 + logx 64 =12​

Answer 1

Answer:

x = 2

Step-by-step explanation:

$log_{x}(4) + log_{x}(16) + log_{x}(64) = 12 \\ = > log_{x}( {2}^{2} \times {2}^{4} \times {2}^{6} ) = 12 \\ = > log_{x}( {2}^{12} ) = 12 \\ = > 12 log_{x}(2) = 12 \\ = > log_{x}(2) = 1 \\ = > x = 2$

IDENTITY USED:

log_a(b) + log_a(c) = log_a(b×c)

log_a(bⁿ) = nlog_a(b)

HOPE IT HELPS :D