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16x < 1284
= ( 24 )x < ( 27)4
= 24x < 228
= 4x < 28
=> x < 7
Vậy x ∈ { 0; 1; 2; 3; 4; 5; 6 }.
\(16^x< 128^4\)
=> \(\left[2^4\right]^x< \left[2^7\right]^4\)
=> \(2^{4x}< 2^{28}\)
=> 4x < 28
=> x < 7
Đến đây tìm x được rồi
\(\left[3x^2-5\right]+3^4+6^0=5^3\)
=> \(\left[3x^2-5\right]=5^3-6^0-3^4=43\)
=> \(3x^2-5=43\)
=> \(3x^2=48\)
=> \(x^2=16\)
=> \(x=\pm4\)
\(3x+2x\left[2^3\cdot5-3^2\cdot4\right]+5^2=4^4\)
=> \(3x+2x\left[8\cdot5-9\cdot4\right]+25=256\)
=> \(3x+2x\cdot4+25=256\)
=> \(3x+2x\cdot4=231\)
Đến đây tìm x
**(3x - 16 ) . 343 = 2.74
(3x - 16 ) . 73 = 2.7 . 73
3x - 16 = 14
3x = 14 + 16 = 30
X =10
** 53 - 7x = 57 : 55
53 - 7x = 52 = 25
7x = 53 - 25 = 28
x = 28 : 7 = 4
\(\left[x^2\right]^4\cdot16=2^{20}\)
\(\Leftrightarrow x^8\cdot16=2^{20}\)
\(\Leftrightarrow x^8\cdot2^4=2^{20}\)
\(\Leftrightarrow x^8=2^{20}:2^4\)
\(\Leftrightarrow x^8=2^{16}\)
\(\Leftrightarrow x^8=\left[2^2\right]^8\)
\(\Leftrightarrow x^8=4^8\Leftrightarrow x=4\)
\(2^{x+3}+2^x=144\)
\(\Leftrightarrow2^x\cdot2^3+2^x=144\)
\(\Leftrightarrow2^x\left[2^3+1\right]=144\)
\(\Leftrightarrow2^x\cdot9=144\)
\(\Leftrightarrow2^x=16\Leftrightarrow x=4\)
Bài cuối tt
X^8=2^16 =>X=4
2^X(2^3+1)=144
=>2^X=16
=>X=4
3^3X(2^2-1)=9^5
=>3^3X=3^9
=>X=3
a. => \(2^{6+x}=2^{10}\)
=> 6+x=10
=> x=10-6
Vậy x=4.
b. => \(7^{3x-1}:7^2=7^6\)
=> 73x-1-2=76
=> 73x-3=76
=> 3x-3=6
=> 3x=6+3
=> 3x=9
Vậy x=3.
c. =>\(7^{5x-1}-25=24\)
=>75x-1=24+25
=>75x-1=49
=>75x-1=72
=>5x-1=2
=>5x=3
Vậy x=\(\frac{3}{5}\).
d. => \(10^{x-3}=10^0\)
=>x-3=0
Vậy x=3.
e. => 2x=10
=> x=10:2
Vậy x=5.
\(a,x:\left(\frac{-1}{3}\right)^2=\frac{1}{3}\)
\(x:\left(\frac{1}{3}\right)^2=\frac{1}{3}\)
\(x=\frac{1}{3}\times\left(\frac{1}{3}\right)^2=\frac{1}{27}\)
\(b,\left(\frac{4}{5}\right)^5.x=\left(\frac{4}{5}\right)^7\)
\(x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)
\(x=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)
\(c,\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{-1}{4}\)
\(d,\left(3x+1\right)^3=-27\)
\(\left(3x+1\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+1=-3\)
\(3x=-4\)
\(x=\frac{-4}{3}\)
\(\left(3x-5\right)^2=16\)
\(\left(3x-5\right)^2=4^2\) hoặc \(\left(3x-5\right)^2=\left(-4\right)^2\)
\(3x-5=4\) hoặc \(3x-5=-4\)
\(3x=4+5\) hoặc \(3x=-4+5\)
\(3x=9\) hoặc \(3x=1\)
\(x=9:3\) hoặc \(x=1:3\)
\(x=3\) hoặc \(x=\dfrac{1}{3}\)
Vậy \(x=3\) hoặc \(x=\dfrac{1}{3}\)
[3x-5]2=16
[3x-5]2=42
3x-5=4
3x=4+5
3x=9
x=9:3
x=3