Solve the following equation with a rational exponent:  5x43=80 .(1 point)Responses40.3240.3288161612

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2 hours ago

To solve the equation \( 5x^{43} = 80 \) for \( x \), we first isolate \( x^{43} \).

  1. Divide both sides by 5:

\[ x^{43} = \frac{80}{5} \]

\[ x^{43} = 16 \]

  1. Now, we take the 43rd root of both sides. This can be expressed using a rational exponent:

\[ x = 16^{\frac{1}{43}} \]

  1. To simplify \( 16 \) as a power of \( 2 \) (since \( 16 = 2^4 \)):

\[ x = (2^4)^{\frac{1}{43}} = 2^{\frac{4}{43}} \]

Now we can approximate \( 2^{\frac{4}{43}} \):

  • To find \( 2^{\frac{4}{43}} \) we can either use a calculator or compute it approximately.

Using a calculator:

\[ 2^{\frac{4}{43}} \approx 1.037 \]

The options provided are 40.32, 40.328, 816, 161, and 12, none of these values are close to a reasonable approximation of \( 2^{4/43} \). Therefore, check if considering \( 2^4 \) could offer a distinct solution:

The correction should be:

\[ x = 16^{\frac{1}{43}} \text{ or } 2^{\frac{4}{43}} \text{ which doesn't appear in the choices } \]

In conclusion, compute the rough values of these computations with better decimal accuracy or check rounding approaches:

Thus while numerical checking there isn’t proper correspondence to provided options.

Ensure methods for rationalizing \( 2^{\frac{4}{43}} \) computational checks, or \( 80/5 = 16\) rational arithmetic confirms back to check other mathematical methods applied.

Final output of the potential value \( x \) remains approximately rational via outputs remain:

\[ x \approx 1.037 \text{ objective. Check against choices remains absent } \]

If matches require rational equivalents, there may be further context or purpose highlighting methodology in discrepancy of options through designs. Thus, resolving full check however underlying would yield \( x \approx 1.037 \) remains merely for discrepancy not matching offered selections.