What is the power of 12 (15 with 60 as the base of 1-log) divided by 2 (5 with 60 as the base of 1-log)

What is the power of 12 (15 with 60 as the base of 1-log) divided by 2 (5 with 60 as the base of 1-log)

12^{[1-log(60)15]/2[1-log(60)5]}=12^{[log(60)60-log(60)15]/2[log(60)60-log(60)5]}=12^{[log(60)60/15]/2[log(60)60/5]}=12^{[log(60)4]/[log(60)12²]=12^{[log(12)4]/log(12)60]/[log(12)12²/log(12)...

Given that a > 0, a is not equal to 1, 1 + log (4-A ^ x) > = log (a ^ x-1) with (1 / 4) base

To ensure that both sides of the inequality are meaningful, we need 4-A ^ x > 0, a ^ X-1 > 0, that is, 1 = - log (a ^ x-1) log (4) log (4) + log (a ^ x-1) > = 2log (4-A ^ x) log (4) log (4a ^ x-4) > = log [(4-A ^ x) square] with (4) base. Let y = a ^ x, 4y-4 > = (...)

If f (x) = log is based on a, X (a > 0, a is not equal to 1) satisfies f (9) = 2, then f ^ - 1 (log is based on 9, 2) =?!
Thank you for your help!

According to the meaning of the title
Log (a) 9 = 2, so a = 3
That is, f (x) = log (3) X
Let log (3) x = log (9) 2
The solution is x = √ 2
So f ^ - 1 [log (9) 2] = √ 2

What is 2 ^ 1 + log (2) 5

2^(1+log(2)5)
=2^1 * 2 ^log(2)5)
=2 * 5
=10

What does log √ 2 (2) + log (4) (2) equal

A: the formula for changing the bottom
log√2(2)+log4(2)
=log2(2)/log2(√2)+log2(2)/log2(4)
=1/(1/2)+1/2
=2+1/2
=5/2

log(2)3×log(3)4×log(5)6…… How much does × log (2008) 2009 equal? Seeking process and answer, solving the problem of fixed praise!

log(2)3*log(3)4*log(4)5*…… *log(2008)2009
=lg3/lg2xlg4/lg3.lg2009/lg2008
=lg2009/lg2=log2(2009)

Given that log is based on 2, the true number is 3 = a, log is based on 2, and the true number is 5 = B, then log is based on 2, and the true number is 9 / 5

Log is based on 2 and the true number is 9 / 5
=Log is based on 2, the true number is 9, minus log is based on 2, the true number is 5
=Double log is based on 2, true number is 3, minus log is based on 2, true number is 5
=2a-b

Let a = 30.5, B = log32, C = cos23 π, then the size relation of a, B, C is______ ．

A = 30.5 ＞ 30 = 1, 0 = log31 ＜ B = log32 ＜ log33 = 1, C = cos23 π = - cos π 3 ＜ 0, C ＜ B ＜ a

Let a = 0.5 ^ - 0.5, B = log0.30.4, C = Cos2 π / 3, judge the size of a, B, C

a=0.5^-0.5,
a=1/√0.5>1
b=log0.3 0.4 ,
0c

Given a = log (2) 3 + log (2) √ 3, B = log (2) 9-log (2) √ 3, C = log (3) 2, then the size relationship of a, B, C is (detailed process,)
A.a=b＜c B.a=b＞c C.a＜b＜c D.a＞b＞c

a=log[2]3√3＞1
b=log[2]（9/√3）=log[2]3√3=a
c=log[3]2＜1
So C < B = a