# Solving $(-2.1)^x = -9.261$ [closed]

Here a given equation:

$$(-2.1)^x = -9.261$$

Here we have a problem when we begin to solve the equation. The problem is that I cant take the logarithm function to the both sides, because $$x <1$$, I tried to multiple both sides by $$-1$$, But it's useless because you still have a minus sign.

## closed as unclear what you're asking by Xander Henderson, RRL, José Carlos Santos, John Omielan, Parcly TaxelMar 26 at 3:26

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• What is $x$ supposed to be? Integer, real, complex? If one of the latter: how do you define exponentation of a negative number by a real one, i.e. how would you define $(-2.1)^\pi$? – mrtaurho Mar 24 at 17:16
• Without additional context, it is going to be difficult to provide you with a reasonable answer. What is your definition of the exponential function $a^x$ when $a < 0$? Why have you assumed that $x < 1$? Are you familiar with the complex logarithm, or are you working in a situation where the only variables are real? Please edit your question to provide this context. – Xander Henderson Mar 24 at 17:16
• It gives complex solutions $$1.105655862- 0.4473796622\,i$$ – Dr. Sonnhard Graubner Mar 24 at 17:19
• Hint: 9261 is a perfect... something. Use guess and check. – Sean Roberson Mar 24 at 17:19
• Here is $X = 3$ , I Know that $X$ should not be a fraction. – Mohammad Alshareef Mar 24 at 17:21

If we assume integer result, we know that $$x$$ must be odd since left hand side of equation is negative.
So $$x=2k+1$$ That means $$(-2.1)^{2k+1}=-9.261$$

$$(-2.1)^{2k}(-2.1)^1=-9.261$$

$$(-2.1)^{2k}=4.41$$

$$[(-2.1)^2]^k=4.41$$

$$(4.41)^k=4.41$$

So $$k=1$$

And $$x=2(1)+1$$

• This is an excellent way of thinking , Thank you very much. – Mohammad Alshareef Mar 24 at 17:46

This appears to be a precalculus level problem where the student is expected to notice that

1. $$-2.1=-\dfrac{3\cdot7}{10}$$
2. $$-9.261=-\dfrac{9261}{1000}=-\dfrac{3^3\cdot7^3}{10^3}=\left(-\dfrac{3\cdot7}{10}\right)^3$$
3. $$(-2.1)^3=-9.261$$

So $$x=3$$