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How to solve this for $x$:

$$\log_x(x^3+1)\cdot\log_{x+1}(x)>2$$

I have tried to get the same exponent by getting the second multiplier to reciprocal and tried to simplify $(x^3+1)$.

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    $\begingroup$ Hint: $x^3+1=(x+1)(x^2-x+1)$. $\endgroup$ – vadim123 Jun 29 '15 at 18:00
  • $\begingroup$ Recall that $log_x(x^3+1)=a$ means that $x^a=x^3+1$. $\endgroup$ – Chuks Jun 29 '15 at 18:04
  • $\begingroup$ I will try something with that thank you. $\endgroup$ – zivce Jun 29 '15 at 18:11
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We have that

$$\begin{align} \log_{x+1}x&=\frac{\log_xx}{\log_x(x+1)}\\\\ &=\frac{1}{\log_x(x+1)} \tag 1 \end{align}$$

and

$$\log_x(x^3+1)=\log_x (x+1)+\log_x(x^2-x+1) \tag 2$$

Using $(1)$ and $(2)$ reveals that

$$\begin{align} \log_x(x^3+1)\log_{x+1}x&=\left(\log_x (x+1)+\log_x(x^2-x+1)\right)\frac{1}{\log_x(x+1)}\\\\ &=1+\frac{\log_x(x^2-x+1)}{\log_x(x+1)} \tag 3 \end{align}$$

We note that if the right-hand side of $(3)$ is to be greater than $2$, we must have $\frac{\log_x(x^2-x+1)}{\log_x(x+1)}>1\implies \frac{x^2-x+1}{x+1}>1\implies x(x-2)>0\implies x>2 \,\,\text{for real-valued solutions}$

Thus, we have that

$$\bbox[5px,border:2px solid #C0A000]{\log_x(x^3+1)\log_{x+1}x>2\,\,\text{for}\,\,x>2}$$

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Keep in mind $(\log_a b)(\log_c a) = \log_c b$.

Setting $a = x, b = x^3 + 1$, and $c = x + 1$, we have
$\log_x (x^3 + 1)\cdot \log_{x+1}x = \log_{x + 1}(x^3 + 1)$.

So your inequality is equivalent to
$\log_{x + 1}(x^3 + 1) > 2 \\ x^3 + 1 > (x + 1)^2 \\ x^3 + 1 > x^2 + 2x + 1 \\ x^3 - x^2 - 2x > 0 \\ x(x + 1)(x - 2) > 0$

Then $-1 < x < 0$ or $x > 2$. One case is ruled out.

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  • $\begingroup$ Good to see you back! $\endgroup$ – Mark Viola Jun 29 '15 at 19:28

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