ahhadyhccjdhratssgxg $f,~2^x.4=\frac1{64}$,$q,~3^x+3=16$,$h,~2^{x-1}-5=11$,$i,~(\frac12)^{x-5}=4.$,$j,~(\frac13)^{x^2+3x-3}=\frac13.$,$h,~2^{3x-1}=\frac1{2^{x-1}}$,$l,~2.e^{2x}=5$,$m,~3^{x+1}=\frac1{3^{1-2x}}$

Question

Accepted Answer

a) $2^x \cdot 4 = \frac{1}{64}$
$2^x \cdot 2^2 = 2^{-6}$
$2^{x+2} = 2^{-6}$
$x + 2 = -6$
$x = -8$

b) $3^x + 3 = 16$
$3^x = 13$
$x = \log_3(13)$

c) $2^{x-1} - 5 = 11$
$2^{x-1} = 16$
$2^{x-1} = 2^4$
$x - 1 = 4$
$x = 5$

d) $\left(\frac{1}{2}\right)^{x-5} = 4$
$\left(2^{-1}\right)^{x-5} = 2^2$
$2^{-(x-5)} = 2^2$
$-(x-5) = 2$
$x - 5 = -2$
$x = 3$

e) $\left(\frac{1}{3}\right)^{x^2 + 3x - 3} = \frac{1}{3}$
$\left(3^{-1}\right)^{x^2 + 3x - 3} = 3^{-1}$
$3^{-(x^2 + 3x - 3)} = 3^{-1}$
$-(x^2 + 3x - 3) = -1$
$x^2 + 3x - 3 = 1$
$x^2 + 3x - 4 = 0$
$(x + 4)(x - 1) = 0$
$x = -4$ hoặc $x = 1$

f) $2^{3x-1} = \frac{1}{2^{x-1}}$
$2^{3x-1} = 2^{-(x-1)}$
$3x - 1 = -(x - 1)$
$3x - 1 = -x + 1$
$4x = 2$
$x = \frac{1}{2}$

g) $2 \cdot e^{2x} = 5$
$e^{2x} = \frac{5}{2}$
$2x = \ln\left(\frac{5}{2}\right)$
$x = \frac{1}{2} \ln\left(\frac{5}{2}\right)$

h) $3^{x+1} = \frac{1}{3^{1-2x}}$
$3^{x+1} = 3^{-(1-2x)}$
$x + 1 = -(1 - 2x)$
$x + 1 = -1 + 2x$
$1 + 1 = 2x - x$
$2 = x$
$x = 2$

Đáp số:
a) $x = -8$
b) $x = \log_3(13)$
c) $x = 5$
d) $x = 3$
e) $x = -4$ hoặc $x = 1$
f) $x = \frac{1}{2}$
g) $x = \frac{1}{2} \ln\left(\frac{5}{2}\right)$
h) $x = 2$