How come $a^{x^3}>(1/a)^{x^2-1}$ has the same solutions as $x^3+x^2-1>0$ if $a > 1$?

Question

How come $$a^{x^3}>(1/a)^{x^2-1}$$ has the same solutions as $$x^3+x^2-1>0$$

if

$$a > 1$$

Also, what type of math is needed to know the answer to this?

Answer 1

Notice that $$(1/a)^{x^2-1}=a^{-(x^2-1)}$$ so your inequality becomes $$a^{x^3}>a^{-(x^2-1)}$$ Taking $\log_a$ on both sides assuming $a>0$, $${x^3}>{-(x^2-1)}$$ Rearranging, $$x^3+x^2-1>0$$

Answer 2

HINT

Note that

$$a^{x^3}>(1/a)^{x^2-1}\iff a^{x^3}>(a)^{1-x^2} $$

then consider that

• for $0<a<1$, $a^x$ is decreasing
• for $a=1$, $a^x$ is constant
• for $a>1$, $a^x$ is increasing and then $a^y>a^z\iff y>z$

Answer 3

Hint: Taking the logarithm on both sides we get $$x^3\ln(a)>(x^2-1)\ln\left(\frac{1}{a}\right)$$ and $$\ln\left(\frac{1}{a}\right)=\ln(1)-\ln(a)$$ So we get $$x^3\ln(a)>(x^2-1)(-\ln(a))$$ Can you finish? And we get $$(x^3+x^2-1)\ln(a)>0$$ if $\ln(a)>0$ this means $a>1$ then we get $x^3+x^2-1>0$

Answer 4

$$a^{x^3}>(1/a)^{x^2-1}$$

Multiply both sides by $a^{x^2-1}$ to get:

$$a^{x^3+x^2-1}>1$$ For this to be true, the exponent of $a$ has to be $>0$, that is, $$x^3+x^2-1>0$$

Answer 5

Since $a\gt0$, we have

$$a^{x^3}\gt (1/a)^{x^2-1}\iff a^{x^3+x^2-1}\gt1$$

and since $a\gt1$, we have

$$a^u\gt1\iff u\gt0$$

Answer 6

A bit of nitpicking:

Let $a>1.$

User Karn has shown that a solution of

1) $a^{x^3} >(1/a)^{x^2-1}$ , i.e a $x$ that satisfies the inequality (1) is a solution of 2) $x^3+x^2-1>0$, i.e satisfies the inequality (2).

Remains to be shown that if a $x$

satisfies 2), i.e. $x^3+x^2-1 >0$, then this $x$ also satisfies 1).

$\Leftarrow:$

$F(x):=a^x$ is strictly increasing, hence:

$x^3 >-x^2+1$ implies $a^{x^3} > a^{-(x^2-1)} =(1/a)^{x^2-1}$.

Done.