Giải phương trình: \(\log_3\left(4^x-1\right)=\log_4\left(3^x+1\right)\)
giải các bất phương trình sau
a) \(log\left(x-5\right)< 2\)
b) \(log_2\left(2x-3\right)>4\)
c) \(log_3\left(2x+5\right)\le3\)
d) \(log_4\left(4x-5\right)\ge2\)
e) \(log_3\left(1-3x\right)>3\)
a: \(log\left(x-5\right)< 2\)
=>\(\left\{{}\begin{matrix}x-5>0\\log\left(x-5\right)< log4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-5>0\\x-5< 4\end{matrix}\right.\Leftrightarrow5< x< 9\)
b: \(log_2\left(2x-3\right)>4\)
=>\(log_2\left(2x-3\right)>log_216\)
=>\(\left\{{}\begin{matrix}2x-3>0\\2x-3>16\end{matrix}\right.\)
=>2x-3>16
=>2x>19
=>\(x>\dfrac{19}{2}\)
c: \(log_3\left(2x+5\right)< =3\)
=>\(log_3\left(2x+5\right)< =log_327\)
=>\(\left\{{}\begin{matrix}2x+5>0\\2x+5< =27\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-\dfrac{5}{2}\\x< =11\end{matrix}\right.\)
=>\(-\dfrac{5}{2}< x< =11\)
d: \(log_4\left(4x-5\right)>=2\)
=>\(log_4\left(4x-5\right)>=log_416\)
=>4x-5>=16 và 4x-5>0
=>4x>=21 và 4x>5
=>4x>=21
=>\(x>=\dfrac{21}{4}\)
e: \(log_3\left(1-3x\right)>3\)
=>\(log_3\left(1-3x\right)>log_327\)
=>\(\left\{{}\begin{matrix}1-3x>0\\1-3x>27\end{matrix}\right.\)
=>1-3x>27
=>\(-3x>26\)
=>\(x< -\dfrac{26}{3}\)
↑\(\log_2X+\log_3\left(X+1\right)< \log_4\left(X+2\right)+\log_5\left(X+3\right)\)
giải phương trình
a) \(logx=2\)
b) \(log_45x=log_4\left(2x-3\right)\)
c) \(log7x=4\)
d) \(ln\left(x^2+5\right)=ln6x\)
a.
ĐKXĐ: \(x>0\)
\(logx=2\Rightarrow x=10^2=100\)
b.
ĐKXĐ: \(x>\dfrac{3}{2}\)
\(log_45x=log_4\left(2x-3\right)\Rightarrow5x=2x-3\)
\(\Rightarrow x=-1\) (ktm ĐKXĐ)
Vậy pt vô nghiệm
c.
ĐKXĐ: \(x>0\)
\(log7x=4\Rightarrow7x=10^4\)
\(\Rightarrow x=\dfrac{10^4}{7}\)
d.
ĐKXĐ: \(x>0\)
\(ln\left(x^2+5\right)=ln\left(6x\right)\Rightarrow x^2+5=6x\)
\(\Leftrightarrow x^2-6x+5=0\Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Giải phương trình
\(\log_2x+\log_3\left(x+1\right)=\log_4\left(x+2\right)+\log_5\left(x+3\right)\)
Điều kiện x>0. Nhận thấy x=2 là nghiệm.
Nếu x>2 thì
\(\frac{x}{2}>\frac{x+2}{4}>1\); \(\frac{x+1}{3}>\frac{x+3}{5}>1\)
Suy ra
\(\log_2\frac{x}{2}>\log_2\frac{x+2}{4}>\log_4\frac{x+2}{4}\)hay :\(\log_2x>\log_2\left(x+2\right)\)
\(\log_3\frac{x+1}{3}>\log_3\frac{x+3}{5}>\log_5\frac{x+3}{5}\) hay \(\log_3\left(x+1\right)>\log_5\left(x+3\right)\)
Suy ra vế trái < vế phải, phương trình vô nghiệm.
Đáp số x=2
Bài 1:
Giari PT: \(\frac{1}{2}log_{\sqrt{2}}\left(x+3\right)+\frac{1}{4}log_4\left(x-1\right)^8=3log_8\left(4x\right)\)
Bài 2:
Tìm m để PT sau có nghiệm: \(x\in\left[0;1+\sqrt{3}\right]\):
\(m\left(\sqrt{x^2-2x+2}+1\right)+x\left(2-x\right)\le0\)(2)
Bài 3:
Giari HPT: \(\hept{\begin{cases}x^4-4x^2+y^2-6y+9=0\\x^2y+x^2+2y-22=0\end{cases}}\)(2)
P/s: Mình không cần gấp,cuối tuần mình mới nộp. Cac bạn gắng giúp mình nha!
Giải các phương trình sau :
a) \(5^{\cos\left(3x+\dfrac{\pi}{6}\right)}=1\)
b) \(6.4^x-13.6^x+6.9^x=0\)
c) \(7^{x^2}.5^{2x}=7\)
d) \(\log_4\left(x+2\right)\log_x2=1\)
e) \(\dfrac{\log_3x}{\log_93x}=\dfrac{\log_{27}9x}{\log_{81}27x}\)
g) \(\log_3x+\log_4\left(2x-2\right)=2\)
\(\log_4\left(\log_2x\right)+\log_2\left(\log_4x\right)\ge2\)
ĐKXĐ: \(x>1\)
\(\Leftrightarrow\frac{1}{2}log_2\left(log_2x\right)+log_2\left(\frac{1}{2}log_2x\right)\ge2\)
\(\Leftrightarrow log_2\left(\frac{1}{2}log_2x.\sqrt{log_2x}\right)\ge2\)
\(\Leftrightarrow\frac{1}{2}\sqrt{log_2^3x}\ge4\Leftrightarrow\sqrt{log^3_2x}\ge8\)
\(\Leftrightarrow log_2^3x\ge64\Leftrightarrow log_2x\ge4\)
\(\Rightarrow x\ge16\)
Tính giá trị biểu thức :
\(A=\log_{2013}\left\{\log_4\left(\log_2256\right)-\log_{0,25}\left[\log_9\left(\log_464\right)\right]\right\}\)
\(A=\log_{2013}\left\{\log_4\left(\log_2256\right)-\log_{0,25}\left[\log_9\left(\log_464\right)\right]\right\}=\log_{2013}\left\{\log_4\left(\log_22^8\right)-\log_{0,25}\left[\log_9\left(\log_44^3\right)\right]\right\}\)
\(=\log_{2013}\left\{\log_48-\log_{0,25}\log_93\right\}=\log_{2013}\left\{\log_{2^2}2^2-\log_{\left(\frac{1}{2}\right)^2}\frac{1}{2}\right\}\)
\(=\log_{2013}\left(\frac{3}{2}-\frac{1}{2}\right)=\log_{2013}1=0\)
