a: \(5^{x}\cdot\left(5^3\right)^2=625\)

=>\(5^{x}=\frac{5^4}{5^6}=5^{-2}\)

=>x=-2

b: \(\left(\frac{12}{15}\right)^{x}=\left(\frac53\right)^{-5}-\left(-\frac35\right)^4\)

=>\(\left(\frac45\right)^{x}=\left(\frac35\right)^5-\left(\frac35\right)^4=\left(\frac35\right)^4\cdot\left(\frac35-1\right)=\left(\frac35\right)^4\cdot\frac{-2}{5}=\frac{-2\cdot3^4}{5^5}\)

=>\(x=\log_{0,8}\left(-2\cdot\frac{3^4}{5^5}\right)\)

c: \(\left(-\frac34\right)^{3x-1}=\frac{256}{81}\)

=>\(\left(-\frac34\right)^{3x-1}=\left(-\frac34\right)^{-4}\)

=>3x-1=-4

=>3x=-3

=>x=-1

d: \(172x^2-7^9:98^3=2^{-3}\)

=>\(172x^2=\frac18+\frac{7^9}{7^6\cdot2^3}=\frac18+\frac{7^3}{2^3}=\frac{1+343}{8}=\frac{344}{8}\)

=>\(x^2=\frac{344}{8}:172=\frac{344}{8\cdot172}=\frac28=\frac14\)

=>\(\left[\begin{array}{l}x=\frac12\\ x=-\frac12\end{array}\right.\)

 dề bài đâu mà tìm?

ủa mù à

\(\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)

\(\Rightarrow\left(\dfrac{-3}{4}\right)^{3x-1}=\left(\dfrac{4}{3}\right)^4\)

Xem lại đề

\(\left(-\dfrac{3}{4}\right)3x-1=\dfrac{256}{81}\)

\(-\dfrac{9}{4}\cdot x=\dfrac{337}{81}\)

\(x=\dfrac{337}{81}\cdot-\dfrac{4}{9}\)

\(x=-\dfrac{1348}{729}\)

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây

Lời giải:

a.

$(\frac{-1}{3})^3.x=\frac{1}{81}=(\frac{-1}{3})^4$
$\Rightarrow x=(\frac{-1}{3})^4: (\frac{-1}{3})^3=\frac{-1}{3}$
b.

$2^2.16> 2^x> 4^2$
$\Rightarrow 2^2.2^4> 2^x> (2^2)^2$

$\Rightarrow 2^6> 2^x> 2^4$

$\Rightarrow 6> x> 4$

$\Rightarrow x=5$ (với điều kiện $x$ là số tự nhiên nhé)

c.

$9.27< 3^x< 243$

$3.3^3< 3^x< 3^5$

$\Rightarrow 3^4< 3^x< 3^5$

$\Rightarrow 4< x< 5$

Với $x$ là stn thì không có số nào thỏa mãn.

a: \(\Leftrightarrow\left(\dfrac{12}{25}\right)^x=\dfrac{9}{25}-\dfrac{81}{625}=\dfrac{144}{625}\)

=>x=2

b: =>3x-1=-4

=>3x=-3

hay x=-1

a: \(2^{x^2-1}=256\)

=>\(2^{x^2-1}=2^8\)

=>\(x^2-1=8\)

=>\(x^2=9\)

=>\(x\in\left\{3;-3\right\}\)

b: \(3^{x^2+3x}=81\)

=>\(3^{x^2+3x}=3^4\)

=>\(x^2+3x=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

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

c: \(2^{x^2-5x}=64\)

=>\(2^{x^2-5x}=2^6\)

=>\(x^2-5x=6\)

=>\(x^2-5x-6=0\)

=>(x-6)(x+1)=0

=>\(\left[{}\begin{matrix}x-6=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}\right)^x=243\)

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

=>x=-5

e: \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)

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

=>-x-5=2x+1

=>-3x=6

=>x=-2

Ai giúp mình với, ai có câu trả lời chính xác mình sẽ kết bạn với nick đó

a, \(\left(1+3x\right)^4=256\)

\(\left(1+3x\right)^4=4^4\)

\(\Rightarrow1+3x=4\)

\(\Rightarrow3x=3\)

\(x=1\)

b, \(A=\dfrac{3^{15}.29+3^{15}.88}{3^{13}.81}\)

\(A=\dfrac{3^{15}\left(29+88\right)}{3^{13}.81}\)

\(A=\dfrac{3^{15}.117}{3^{13}.3^4}\)

\(A=\dfrac{3^{15}.3^2.13}{3^{17}}\)

\(A=\dfrac{3^{17}.13}{3^{17}}=13\)