Pt mũ đưa về cùng cơ số

Question

$1) 2^{x+4}+2^{x+2}=5^{x+1}+3.5^x$
^{$2) 3^{x-1}=18^{2x}.2^{-2x}.3^{x+1}$}^{$3) 2^x.3^{x+1}=(\sqrt3)^{x+2}$}

score 1 · Answer 1

2x+4+2x+2=5x+1+3.5x

$\Leftrightarrow 16.2^x + 4.2^x = 5.5^x + 3.5^x = 8.5^x$

$\Leftrightarrow 20.2^x = 8.5^x$

$\Leftrightarrow 5.2^x = 2.5^x$

$\Leftrightarrow 2^{x-1} = 5^{x-1}$

$\Leftrightarrow x-1 = (x-1)\log_2 5$

$\Leftrightarrow x = 1$

score 1 · Answer 2

3x−1=182x2−2x3x+1

$\Leftrightarrow 3^{x-1} = 2^{2x} . 3^{6x} . 2^{-2x} . 3^{x+1} = 3^{7x +1}$

$\Leftrightarrow 3^{7x +1 - x + 1} = 1$

$\Leftrightarrow 3^{6x +2} = 1$

$\Leftrightarrow 6x +2 = 0 \Rightarrow x = -\dfrac{1}{3}$

uh nhầm 18^(2x) = [2.3^2]^(2x) = 2^(2x) . 3^2(2x) vậy là 3^(4x) – Dép Lê Con Nhà Quê 06-08-13 04:47 PM

hình như bạn tách nhầm chỗ 18 thì phải – baongoc.kuty.3 06-08-13 10:21 AM

score 1 · Answer 3

2x.3x+1=3√x+2

$\Leftrightarrow 2^x .3^{x+1} = 3^{\frac{x+2}{2}}$

$\Leftrightarrow 2^x = 3^{\frac{x+2}{2} - x - 1} = 3^{-\frac{x}{2}}$

$\Leftrightarrow x = -\dfrac{x}{2}\log_2 3$

$\Leftrightarrow x = 0$