How to prove log (2) (x) + x < 4 ^ (x-1), (x > = 2)

How to prove log (2) (x) + x < 4 ^ (x-1), (x > = 2)

Let f (x) = 4 ^ (x-1) - x-logx, x > = 2, then
f'(x)=4^(x-1)*ln4-1-1/(xln2)
>=f'(2)=4ln4-1-1/ln4>0,
If f (x) is an increasing function,
∴f(x)>=f(2)=4-2-1=1>0,
The proposition holds

Find the range of log (1 / 2) cos (x + π / 6) ≤ 1 / 2, X

log（1/2）cos(x+π/6)≤1/2=log(1/2)[(1/2)^1/2]=log(1/2)(√2/2)
Log (1 / 2) x decreasing
So cos (x + π / 6) ≥ √ 2 / 2 = cos (2k π ± π / 4)
So 2K π - π / 4 ≤ x + π / 6 ≤ 2K π + π / 4
2kπ-5π/12≤x≤2kπ+π/12

log(1/2) (x+1)>log(1/4) (3x+4)
Finding logarithmic inequality

log(1/2)(x+1)>log(1/4)(3x+4)
log(1/2)2(x+1)2>log(1/4)(3x+4)
Y = log (1 / 4) (x) is a decreasing function
So (x + 1) 2

It is known that f (x) = {2 ^ x, (x ≥ 4); f (x + 1) (x)

∵ log2(3)∈(1,2)∴ f(log2(3))=f(log2(3)+1)=f(log2(6))∵ log2(6)∈(2,3)=f(log2(6)+1)=f(log2(12))∵ log2(12)∈(3,4)=f(log2(12)+1)=f(log2(24)∵ log2(24)∈(4,5)=2^(log2(24))=24

If f (x) = 12x (x ≥ 4) f (x + 1) (x < 4), then f (log23) = ()
A. 124B. −238C. 111D. 119

∵ f (x) = 12x (x ≥ 4) f (x + 1) (X & lt; 4), ∵ f (log23) = f (log23 + 1) = f (log26 + 1) = f (log212) = f (log212 + 1) = f (log224) = 12log224 = 124

If a > 0, a ^ 2 / 3 = 4 / 9, log2 / 3 is the logarithm of base 2

A ^ (2 / 3) = 4 / 9 > > > take the reciprocal to get:
Log with (2 / 3) as base 2 log = 3log with a as base 2 log

Find the domain of y = 1 / log2 (x + 1) - 3
Who is right? I'm blinded!

x+1>0
log2(x+1)≠3
therefore
x>-1
x+1≠8====>x≠7
To sum up, x > - 1 and X ≠ 7

Given the function f (x) = log2 (2x-3) + 3 {2 is the base}, find the domain of F (x)?
To specific steps Oh!
O(∩_ Thank you!

Given the function f (x) = log2 (2x-3) + 3 {2 is the base},
2x-3＞0
2x＞3
x＞3/2
The domain of definition is (3 / 2, + ∞)

Ask a math problem. Log (2) 3 * log (3) 4 * log (5) 6 * *log(2008)2009
(2) And so on, under the log, * is a multiplier sign

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

If a = log takes 2 as the bottom 0.9, a = 3 (- 1 / 3) and C = (1 / 3) (1 / 2), then a, B
If a = log takes 2 as the bottom 0.9, a = 3 (- 1 / 3) and C = (1 / 2), then the size relationship of a, B and C is

a