Set a = {x | 2 base (x-3) logarithm > 1}, B = {x | 2 X-1 power > 2}, a is contained in B, find the value range of real number a

Set a = {x | 2 base (x-3) logarithm > 1}, B = {x | 2 X-1 power > 2}, a is contained in B, find the value range of real number a

From loga (x-3) > 1, we get loga (x-3) > loga (a),
Therefore, when a > 1, x-3 > A, x > A + 3, a = {x | x > A + 3},
0

The log of 3E takes three as the base and the logarithm of 5 minus one power

=3e^[log3(5)-log3(3)]
=3e^[log3(5÷3)]
=3e^[log3(5/3)]

2 log5 10 + log5 0.25-3 log2 64 is so anxious!

2log5 10＋log5 0.25－3log2 64
=log5 100+log5 0.25-3×6
=log5(100×0.25)-18
=2-18
=-16

log2(0.2)×log5(8)+10^lg3=______ .

log2(0.2)=lg(0.2)/lg2=-lg5/lg2
log5(8)=log5(2^3)=3log5(2)=3lg2/lg5
So log2 (0.2) × log5 (8) = - 3
10^lg3=3
The original formula = - 3 + 3 = 0
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