Bài 5: Phương trình mũ và phương trình lôgarit

a: \(16^{-x}=8^{2\left(1-x\right)}\)

=>\(2^{-4x}=2^{6\left(1-x\right)}\)

=>6(1-x)=-4x

=>6-6x=-4x

=>-4x+6x=6

=>2x=6

=>x=6/2=3

b: \(3^{\left|3x-4\right|}=9^{2x-2}\)

=>\(3^{\left|3x-4\right|}=3^{4x-4}\)

=>\(\left|3x-4\right|=4x-4\)

=>\(\left\{{}\begin{matrix}4x-4>=0\\\left(4x-4\right)^2=\left(3x-4\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=1\\\left(4x-4-3x+4\right)\left(4x-4+3x-4\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=1\\x\left(7x-8\right)=0\end{matrix}\right.\)

=>\(x=\dfrac{8}{7}\)

c: \(32^{\dfrac{x+5}{x-4}}=0,25\cdot128^{\dfrac{x+17}{x-3}}\)

=>\(2^{\dfrac{5x+25}{x-4}}=2^{-2}\cdot2^{\dfrac{7x+119}{x-3}}\)

=>\(\dfrac{5x+25}{x-4}=-2+\dfrac{7x+119}{x-3}\)

=>\(\dfrac{5x+25}{x-4}=\dfrac{-2x+6+7x+119}{x-3}\)

=>\(\dfrac{5x+25}{x-4}=\dfrac{5x+125}{x-3}\)

=>\(\left(5x+25\right)\left(x-3\right)=\left(5x+125\right)\left(x-4\right)\)

=>\(\left(x+5\right)\left(x-3\right)=\left(x+25\right)\left(x-4\right)\)

=>\(x^2-3x+5x-15=x^2-4x+25x-100\)

=>2x-15=21x-100

=>2x-21x=-100+15

=>-19x=-85

=>\(x=\dfrac{85}{19}\)

\(log_{\sqrt{pq}}\left(\dfrac{q}{\sqrt{p}}\right)=2log_{pq}\left(\dfrac{q}{\sqrt{p}}\right)=2log_{pq}q-2log_{pq}\sqrt{p}\)

\(=\dfrac{2}{log_qpq}-log_{pq}p=\dfrac{2}{log_qp+log_qq}-\dfrac{1}{log_ppq}=\dfrac{2}{\dfrac{1}{\sqrt{5}}+1}-\dfrac{1}{log_pp+log_pq}\)

\(=\dfrac{2}{\dfrac{1}{\sqrt{5}}+1}-\dfrac{1}{1+\sqrt{5}}=\dfrac{11-3\sqrt{5}}{4}\)

\(a^2+b^2=7ab\Leftrightarrow a^2+b^2+2ab=9ab\)

\(\Leftrightarrow\left(a+b\right)^2=9ab\Leftrightarrow\dfrac{\left(a+b\right)^2}{9}=ab\)

\(\Leftrightarrow\left(\dfrac{a+b}{3}\right)^2=ab\)

Lấy logarit cơ số 2 hai vế:

\(log_2\left(\dfrac{a+b}{3}\right)^2=log\left(ab\right)\)

\(\Leftrightarrow2log_2\left(\dfrac{a+b}{3}\right)=log_2a+log_2b\)

d: \(5^{2x}-7^x-5^{2x}\cdot35+7^x\cdot35=0\)

=>\(\left(5^{2x}-7^x\right)-35\left(5^{2x}-7^x\right)=0\)

=>\(\left(5^{2x}-7^x\right)\cdot\left(-34\right)=0\)

=>\(5^{2x}-7^x=0\)

=>\(5^{2x}=7^x\)

=>\(25^x=7^x\)

=>\(\left(\dfrac{7}{25}\right)^x=1\)
=>x=0

g: \(3\cdot25^{x-2}+\left(3x-10\right)\cdot5^{x-2}+3-x=0\)

=>\(3\cdot\left(5^{x-2}\right)^2-5^{x-2}+\left(3x-9\right)\cdot5^{x-2}-\left(x-3\right)=0\)

=>\(5^{x-2}\left(3\cdot5^{x-2}-1\right)+\left(x-3\right)\left(3\cdot5^{x-2}-1\right)=0\)

=>\(\left(5^{x-2}+x-3\right)\left(3\cdot5^{x-2}-1\right)=0\)

=>\(\left[{}\begin{matrix}5^{x-2}+x-3=0\\3\cdot5^{x-2}-1=0\end{matrix}\right.\)

TH1: \(5^{x-2}+x-3=0\)

Đặt \(A\left(x\right)=5^{x-2}+x-3\)

=>\(A\left(x\right)=\dfrac{1}{25}\cdot5^x+x-3\)

=>\(A'\left(x\right)=\dfrac{1}{25}\cdot5^x\cdot ln5+1>0\)

=>A(x) luôn đồng biến trên R

=>A(x)=0 khi x=2

Th2: \(3\cdot5^{x-2}-1=0\)

=>\(5^{x-2}=\dfrac{1}{3}\)

=>\(x-2=log_5\dfrac{1}{3}\)

=>\(x=2+log_5\left(\dfrac{1}{3}\right)\)

h: \(3\cdot4^x+\left(3x-10\right)\cdot2^x+3-x=0\)

=>\(3\cdot4^x-2^x+\left(3x-9\right)\cdot2^x-\left(x-3\right)=0\)

=>\(2^x\left(3\cdot2^x-1\right)+\left(x-3\right)\left(3\cdot2^x-1\right)=0\)

=>\(\left(3\cdot2^x-1\right)\left(2^x+x-3\right)=0\)

TH1: \(3\cdot2^x-1=0\)

=>\(2^x=\dfrac{1}{3}\)

=>\(x=log\left(\dfrac{1}{3}\right)\)

TH2: \(2^x+x-3=0\)

Đặt \(B\left(x\right)=2^x+x-3\)

=>\(B'\left(x\right)=2^x\cdot ln3+1>0\)

=>B(x) luôn đồng biến trên R

=>B(x)=0 khi x=1

c: \(12\cdot3^x+3\cdot15^x-5^{x+1}=20\)

=>\(12\cdot3^x+3\cdot3^x\cdot5^x-5^x\cdot5-20=0\)

=>\(3^x\cdot3\left(5^x+4\right)-5\left(5^x+4\right)=0\)

=>\(\left(3^{x+1}-5\right)\left(5^x+4\right)=0\)

=>\(3^{x+1}-5=0\)

=>\(3^{x+1}=5\)

=>\(x+1=log_35\)

=>\(x=log_35-1\)

f: \(25^x-2\left(3-x\right)\cdot5^x+2x-7=0\)

=>\(\left(5^x\right)^2+5^x\cdot\left(2x-6\right)+2x-7=0\)

=>\(\left(5^x\right)^2+5^x\left(2x-7\right)+5^x+2x-7=0\)

=>\(5^x\left(5^x+2x-7\right)+\left(5^x+2x-7\right)=0\)

=>\(\left(5^x+1\right)\left(5^x+2x-7\right)=0\)

=>\(5^x+2x-7=0\)

Đặt \(A\left(x\right)=5^x+2x-7\)

=>\(A'\left(x\right)=5^x\cdot ln5+2>0\forall x\)

=>A(x) đồng biến trên R

=>A(x)=0 khi và chỉ khi x=1

i: \(9^x+2\left(x-2\right)\cdot3^x+2x-5=0\)

=>\(\left(3^x\right)^2+3^x\left(2x-5\right)+3^x+2x-5=0\)

=>\(\left(3^x+2x-5\right)\left(3^x+1\right)=0\)

=>\(3^x+2x-5=0\)

Đặt \(B\left(x\right)=3^x+2x-5\)

=>\(B'\left(x\right)=3^x\cdot ln3+2>0\)

=>B(x) luôn đồng biến trên R

=>B(x)=0 khi và chỉ khi x=1

\(A=log_m\left(8m\right)=log_mm+log_m8\)

\(=1+log_m8\)

\(=1+\dfrac{1}{log_8m}=1+\dfrac{1}{log_{2^3}m}=1+\dfrac{1}{\dfrac{1}{3}\cdot log_2m}\)

\(=1+\dfrac{1}{\dfrac{1}{3}a}=1+1:\dfrac{a}{3}=1+\dfrac{3}{a}=\dfrac{a+3}{a}\)

=>Chọn A

ĐKXĐ: \(x>0\)

\(log_2^2x-m.log_2x-log_2x+m=0\)

\(\Leftrightarrow log_2x\left(log_2x-m\right)-\left(log_2x-m\right)=0\)

\(\Leftrightarrow\left(log_2x-1\right)\left(log_2x-m\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x_1=2\\x_2=2^m\end{matrix}\right.\)

TH1: \(x_1=x_2^2\Leftrightarrow2=\left(2^m\right)^2=2^{2m}\Rightarrow2m=1\Rightarrow m=\dfrac{1}{2}\)

TH2: \(x_2=x_1^2\Rightarrow2^m=2^2\Rightarrow m=2\)

\(\Rightarrow2+\dfrac{1}{2}=\dfrac{5}{2}\)