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\(1,\\ a,2^x=16=2^4\Rightarrow x=4\\ b,3^{x+1}=9^x=3^{2x}\\ \Rightarrow x+1=2x\Rightarrow x=1\\ c,2^{3x+2}=4^{x+5}=2^{2\left(x+5\right)}\\ \Rightarrow3x+2=2x+10\Rightarrow x=8\\ d,3^{2x-1}=243=3^5\\ \Rightarrow2x-1=5\Rightarrow x=3\\ 2,\\ a,2^{225}=8^{75}< 9^{75}=3^{150}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\\ c,99^{20}=\left(99^2\right)^{10}< \left(99\cdot101\right)^{10}=9999^{10}\\ 3,\\ a,12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}=\left(2\cdot3^2\right)^{16}=18^{16}\\ b,75^{20}=\left(3\cdot5^2\right)^{20}=3^{20}\cdot5^{40}=\left(3^{20}\cdot5^{10}\right)\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}=45^{10}\cdot5^{30}\)
Bài 1:
a) \(\Rightarrow2^x=2^4\Rightarrow x=4\)
b) \(\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)
c) \(\Rightarrow2^{3x+2}=2^{2x+10}\Rightarrow3x+2=2x+10\Rightarrow x=8\)
d) \(\Rightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow x=3\)
Bài 2:
a) \(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)
b) \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\)
c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)
Bài 3:
a) \(12^8.9^{12}=\left(4.3\right)^8.9^{12}=4^8.3^8.9^{12}=2^{16}.9^4.9^{12}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^{16}\)
b) \(75^{20}=\left(75^2\right)^{10}=5625^{10}=\left(45.125\right)^{10}=45^{10}.125^{10}=45^{10}.5^{30}\)
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
a,Tính tổng:S=1+52+54+...+5200
=>52S=52+54+56+...+5202
=>25S-S=24S=5202-1
=>S=\(\frac{5^{202}-1}{24}\)
b,So sánh 230+330+430 và 3.2410
3.24^10=3^11.4^15
4^30=4^15.4^15
hiển nhiên 4^15>3^11
=>3.24^10<<4^30<<<2^30+3^20+4^30
Ta có: 230+330+430>230+230+430=231+230.230
=231(1+229) (1)
Lại có:3.24^10=3^11.2^30 (2)
So sánh (1)và (2): Vì 3^11<4^11=2^22<2^29
và 2^30<2^31
=> 3^11.2^30 <(1+2^29)2^31<2^30+3^30+4^30
a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
b) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
c) \(3^{500}=\left(3^5\right)^{100}=243^{100}\)
\(7^{300}=\left(7^3\right)^{100}=343^{100}>243^{100}\)
\(\Rightarrow3^{500}< 7^{300}\)
a) 32 < 2^n < 128
hay 2^5 < 2^n < 2^7
=> 5 < n < 7
=> n = 6
b) 2.16 \(\ge\)2^n > 4
hay 2^5 \(\ge\)2^n > 2^2
=> 5 \(\ge\)n > 2
=> n \(\in\left\{5;4;3\right\}\)
c) 9.27 \(\le\)3^n \(\le\) 243
hay 3^5 \(\le\)3^n \(\le\) 3^5
=> 5 \(\le\) n \(\le\) 5
=> n = 5
a,32<2^n<128
n sẽ bằng 6 vì khi 2^6=64>32 và 2^6=64 <128 (thỏa mãn điều kiện)
Vậy :n=6
lm tương tự
Bài 5:
a) \(32< 2^n< 128\)
⇔ \(2^5< 2^n< 2^7\)
⇔ \(5< n< 7\)
=> \(n=6\)
Vậy \(n=6.\)
b) Sửa lại đề là \(2.16>2^n>4\)
⇔ \(32>2^n>4\)
⇔ \(2^5>2^n>2^2\)
⇔ \(5>n>2\)
=> \(n=3;n=4\)
Vậy \(n\in\left\{3;4\right\}.\)
c) \(9.27< 3^n< 243\)
⇔ \(243< 3^n< 243\)
⇔ \(3^5< 3^n< 3^5\)
⇔ \(5< n< 5\)
=> \(n\in\varnothing\)
Vậy không tồn tại giá trị nào của \(n.\)
Mình chỉ làm bài 5 thôi nhé.
Chúc bạn học tốt!
5.
a) 32 < 2n < 128
<=> 25 < 2n < 27
<=> 2n = 26
<=> n = 6
b) sai đề
c) 9.27 \(\le\) 3n \(\le\) 243
<=> 35 \(\le\) 3n \(\le\) 35
<=> 3n = 35 <=> n = 5
6.
a) 9920 = (992)10 = 980110
Vì 9801 < 9999 nên 980110 < 999910
hay 9920 < 999910
b) 321 = 3.320 = 3.(32)10 = 3.910
231 = 2.230 = 2.(23)10 = 2.810
Vì 3.910 < 2.810 nên 321 < 231
c) 3.2410 = 3.(23.3)10 = 311.230 = 311.(22)15 = 311.415
Vì 311.415 < 415.415 = 430
nên 3.2410 < 230 + 330 + 430