# Solve this exponential inequality

$${5}^{(x+2)}+{25}^{(x+1)}>750\\ t=5^{(x+1)}\\ t^2+5t-750>0\\ t^2+5t-750=0\\$$

$$a=1, b=5, c= -750\\ D=35+3000=3025 \\ t_1= 25; t_2=-30$$

$t_1=25\Longrightarrow x=1; t_2=-30$ Doesn't have a solution So the solution is $x>1$. Is this correct? I say is bigger than $1$ and not smaller than $1$ because $1$(numer before $t^2$ is bigger than $0$. Is this reason correct or is it because $5>1$ so the exponential function is progressive (the bigger the $x$, the bigger the y)

• $5*5^{x+2}\neq 25^{x+1}$, but $5*5^{x+2}=5^{x+3}$ Sep 11, 2015 at 10:39
• $35+3000 \neq 3025$
– Peđa
Sep 11, 2015 at 10:44

$$5^{(x+2)} = 25 \cdot 5^x$$

$$25^{(x+1)} = 25 \cdot 25^x$$

$$750 = 25 \cdot 30$$

So your inequation may be expressed:

$$5^x + 25^x >30$$

As the derivative of $5^x + 25^x$ is always possitive, there is only one $x$ that makes the equality true:

$$x=1$$

And therefore, the solution of your inequation is $x>1$

• Could you explain why is it x>1 , without using derivative. We haven't learnt them yet Sep 11, 2015 at 10:49
• The idea is that exponential functions (f(x) = a^x) become greater when x increases if a>1, so it is impossible that once you reach a determined value, 30 in your case, you reach it again. Furthermore, as it increases its value with x, the solution is x>=1 instead of x<=1 Sep 11, 2015 at 10:55

Go slower! First note that $25^x=(5^x)^2$ and set $t=5^x$. Then substitute: $$5^{x+2}=5^x\cdot 25=25t, \qquad 25^{x+1}=25^x\cdot25=25t^2$$ Thus the inequality becomes $$25t+25t^2>750$$ and, simplifying, $$t^2+t-30>0$$ or $$(t+6)(t-5)>0$$

$5^{x+2}=25^{\frac{x+2}{2}}$ so: $$25^{\frac{x+2}{2}}+25^{x+1}>750$$ $$25^{\frac{x}{2}}+25^x>30$$ Let's say $25^{\frac{x}{2}}=y$ $$y+y^2=30$$ completing the square: $$(y+0.5)^2>30.25$$ $$y>\sqrt{30.25}-0.5=5$$ remeber that $25^{\frac{x}{2}}=y$ $$25^{\frac{x}{2}}>5$$ square both sides: $$25^x>25$$ this means $x\geq1$