аниқ ҷавоб и керак
Exploring logarithmic solutions
I'm considering the left-hand side (lhs) possibly being \(x^2 + 1\). It seems likely that I should combine it with the right-hand side (rhs) since they're both base powers. This leads me to think the answer might involve logarithms and possibly a ± sign. There’s a lot to consider here, and I want to ensure I'm on the right track!Analyzing the equation
I need to distinguish the expression \(3^{x^2+1}\). If there's ambiguity, I should ask a clarifying question instead of solving right away. However, it seems like I can go ahead with solving this equation rigorously.
Starting with \(5^{x^2} - 3^{x^2+1} = 2(5^{x^2-1} - 3^{x^2-2})\), I expand it and simplify terms. I get down to \(3·5^{x²-1} = 25·3^{x²-2}\). Let's see where this leads!Solving the equation
I’m working through the equation \( \frac{5^{x²-1}}{3^{x²-2}}=25 \). Wait, if I divide by \( 3^{x²-2} \), I have \( 3^{1-(x²-2)} 5^{x²-1}=25 \). I mistakenly approached this earlier. The recalculation shows \( 3 \cdot 5^{x²-1}=25 \cdot 3^{x²-2} \), and dividing gives \( 3/3^{x²-2}=3^{3-x²} \) for the left. This leads to \( x²=3 \), giving \( x=\pm\sqrt{3} \). The calculations are verifying correctly!Solving equations
I’m working on the equation \(3·5^{x²-1}=5²·3^{x²-2}\). First, I rewrite the left and right sides: \(3^{1}5^{(x²-1)}\) and \(5^{2}3^{(x²-2)}\), taking the ratio to get \(3^{3-x²}5^{x²-3}=(3/5)^{3-x²}=1\). This leads me to conclude that \(x²=3\), giving ±√3 as solutions.
Next, I tackle another part involving \(4^{x²+1}=4*4^{x²}\). The roots of \(4t²-9t+2=0\) lead to \(x²=1\) for \(t=2^{x²}\).
I realize it’s useful to make quick checks for \(x=±√3\) and \(x²=1\) to ensure everything checks out. I need to summarize my findings and mention the next steps concisely.Solving the equation
I'm working through the quadratic equation \(t² - 4t - 32 = 0\) and factoring it as \((t - 8)(t + 4)\). That gives me \(t = 8\). Then, moving on to another part, I see \(2^{\sqrt{}} = 2^3\). Since the base 2 is one-to-one, I can set \(\sqrt{} = 3\), resulting in \(x = 6\). It feels like a good step-by-step to reach the solutions!
Краткий ответ
- \(x=\pm1\)
- \(x=6\)
- \(x=\pm\sqrt3\)
Решение
1) \(4^{x^2+1}-9\cdot2^{x^2}+2=0\)
Пусть \(t=2^{x^2}\). Тогда \(4^{x^2+1}=4t^2\):
\[ 4t^2-9t+2=0 \]
\[ (4t-1)(t-2)=0 \]
\[ t=\frac14 \quad \text{или} \quad t=2. \]
Так как \(t=2^{x^2}\ge1\), подходит только \(t=2\):
\[ 2^{x^2}=2 \Rightarrow x^2=1 \Rightarrow \boxed{x=\pm1}. \]
2) \(4^{\sqrt{x+3}}-32=4\cdot2^{\sqrt{x+3}}\)
Пусть \(t=2^{\sqrt{x+3}}>0\). Тогда:
\[ 4^{\sqrt{x+3}}=t^2 \]
\[ t^2-32=4t \]
\[ t^2-4t-32=0 \]
\[ (t-8)(t+4)=0. \]
Так как \(t>0\), \(t=8\):
\[ 2^{\sqrt{x+3}}=8=2^3 \]
\[ \sqrt{x+3}=3 \]
\[ x+3=9 \Rightarrow \boxed{x=6}. \]
3) \(5^{x^2}-3^{x^2+1}=2\left(5^{x^2-1}-3^{x^2-2}\right)\)
Перенесём слагаемые:
\[ 5^{x^2}-2\cdot5^{x^2-1} = 3^{x^2+1}-2\cdot3^{x^2-2}. \]
Вынесем степени:
\[ 3\cdot5^{x^2-1}=25\cdot3^{x^2-2}. \]
Делим на положительное выражение \(25\cdot3^{x^2-2}\):
\[ \left(\frac35\right)^{3-x^2}=1. \]
Основание \(\frac35\neq1\), поэтому показатель равен нулю:
\[ 3-x^2=0 \Rightarrow x^2=3. \]
\[ \boxed{x=\pm\sqrt3}. \]
Проверка для третьего уравнения при \(x^2=3\):
\[ 125-81=2(25-3),\qquad 44=44. \]
