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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.

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ 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 $(-2.1)^\pi$? $\endgroup$ – mrtaurho Mar 24 at 17:16
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    $\begingroup$ 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. $\endgroup$ – Xander Henderson Mar 24 at 17:16
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    $\begingroup$ It gives complex solutions $$ 1.105655862- 0.4473796622\,i$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 24 at 17:19
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    $\begingroup$ Hint: 9261 is a perfect... something. Use guess and check. $\endgroup$ – Sean Roberson Mar 24 at 17:19
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    $\begingroup$ Here is $X = 3$ , I Know that $X$ should not be a fraction. $\endgroup$ – Mohammad Alshareef Mar 24 at 17:21
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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 $x=2(1)+1$

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  • $\begingroup$ This is an excellent way of thinking , Thank you very much. $\endgroup$ – Mohammad Alshareef Mar 24 at 17:46
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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 $x=3$

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