Calculate log4 as the bottom, 27 as the index multiplied by log5 as the bottom, 8 as the index multiplied by log3 as the bottom, and 25 as the index

Calculate log4 as the bottom, 27 as the index multiplied by log5 as the bottom, 8 as the index multiplied by log3 as the bottom, and 25 as the index

log4 27Xlog5 8Xlog3 25
=lg27/lg4 Xlg8/lg5Xlg25/lg3
=3lg3/2lg2X3lg2/lg5X2lg5/lg3
=3/2X3X2
=9

Compare the size of log3 (4) and log4 (5)

Error:
M－N=log(3)[4]－log(4)[5]
=[LG4 / lg3-lg5 / LG4] [bottom changing formula]
=[lg²4－lg3lg5]/[lg3lg4]
Because Lg3 > 0, LG4 > 0
And:
lg3＋lg5≥2√(lg3lg5)
Namely:
lg3lg5≤(1/4)[lg3＋lg5]²=(1/4)lg²15lg3lg5
Then: M-N > 0
Results: log (3) [4] > log (4) [5]

Try to compare the size of log3 1 / 5 and log4 1 / 5

From loga B = 1 / (logba)
log3 1/5=1/log1/5 3
log4 1/5=1/(log1/5 4)
Because 3log1 / 5 4
Namely
1/(log1/5 3)

Known real number a = log45, B = (1 2) 0, C = log30.4, then the size relation of a, B, C is () A. b＜c＜a B. b＜a＜c C. c＜a＜b D. c＜b＜a

∵a=log45＞log44=1，
b=（1
2）0=1，
c=log30.4＜log31=0，
∴c＜b＜a．

How does log3 (4) and log4 (3) compare in size

log3(4)＞log3(3)=1
log4(3)＜log4(4)=1
∴ log3(4)＞log4(3)

If log3 M = log416, then M=

Log 3 M = log 4 16 = 2;
∴3²=；
Then M = 9
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[log4(3)/log4(8)][log3(2)log9(2)] The real numbers are in brackets == I made a mistake [log4 (3) + log8 (3)] [log3 (2) + log9 (2)]

[log4（3）+log8（3）]×[log3（2）+log9（2）]
=[1/2log2（3）+1/3log2（3）]×[log3（2）+1/2log3（2）]
=[5/6log2（3）]×[3/2log3（2）]
=5/6×3/2×log2（3）×log3（2）
=5/6×3/2×1
=5/4

log9^8×log4^27=____ .

log9^8×log4^27
=lg8/lg9×lg27/lg4
=3lg2/2lg3×3lg3/2lg2
=3/2×3/2
=9/4

2^log4(√3+2)^2+3^log9(√3-2)^2=

log4(√3+2)^2=log2(√3+2),log9(√3-2)^2=log3(√3-2).
2^log4(√3+2)^2+3^log9(√3-2)^2
=2^log2(√3+2)+3^log3(√3-2)
=(√3+2)+(√3-2)
=2√3

2^log4(2-√3) ^2+ 3^log9(2+√3)^2 As the title

2^log4(2-√3) ^2+ 3^log9(2+√3)^2
=2^log2(2-√3)+ 3^log3(2+√3)
=2-√3+2+√3
=4