Solving Exponential Equation
We are given the equation $7^{3x} = 216$. Our goal is to find the value of $7^{-x}$ and round it to three decimal places.
Simplifying the Expression
Using the properties of exponents, we can rewrite $7^{3x}$ as $(7^x)^3$. So the equation becomes:
$ (7^x)^3 = 216 $
To find the value of $7^x$, we take the cube root of both sides:
$ 7^x = \sqrt[3]{216} $
We know that $6 \times 6 \times 6 = 216$, so the cube root of 216 is 6:
$ 7^x = 6 $
Calculating the Target Value
The expression we need to evaluate is $7^{-x}$. We can rewrite this using the rule $a^{-n} = \frac{1}{a^n}$:
$ 7^{-x} = \frac{1}{7^x} $
Substitute the value $7^x = 6$ that we found:
$ 7^{-x} = \frac{1}{6} $
Decimal Conversion and Rounding
Finally, we convert the fraction $\frac{1}{6}$ into a decimal and round it to three decimal places:
$ \frac{1}{6} \approx 0.166666... $
Rounding to three decimal places, we get:
$ 0.167 $
This value is between 0.166 and 0.168.
Important Questions from Powers and Exponents
If a real variable $x$ satisfies $3^{x^2} = 27 \times 9^x$, then the value of $\frac{2^{x^2}}{(2^{x})^2}$ is:
- What is the value of $\left(\frac{3^{81}}{27^4}\right)^{1/3}$?
The 12 musical notes are given as C, C#, D, D#, E, F, F#, G, G#, A, A#. Frequency of each note is $ \sqrt[12]{2} $ times the frequency of the previous note. If the frequency of the note C is 130.8 Hz, then the ratio of frequencies of notes F# and C is:
- For positive integers $p$ and $q$, with $\frac{p}{q} \neq 1$, $(\frac{p}{q})^{\frac{p}{q}} = p^{(\frac{p}{q}-1)}$. Then,
- If $x$ satisfies the equation $4^{8x} = 256$, then $x$ is equal to ________.
