Question

\(\log_2(9 - 2^x) = 10^{\log(3-x)}\) solve for \(x\)

• A. 0
• B. 3
• C. both (a) and (b)
• D. 0 and 6

Hint:

Always check domain restrictions for logarithmic functions.

Answer 1

The Correct Option is A

To solve the given equation \(\log_2(9 - 2^x) = 10^{\log(3-x)}\), we will proceed step-by-step with clear logic and calculations.

Step 1: Simplify both sides of the equation

1. Consider the left-hand side of the equation: \(\log_2(9 - 2^x)\).
   • This expression represents the logarithm to the base 2 of the quantity \(9 - 2^x\).
2. On the right-hand side, we have \(10^{\log(3-x)}\).
   • Using the property of logarithms, \(a^{\log_b(c)} = c^{\log_b(a)}\), we can simplify it as: \[ 10^{\log(3-x)} = (3-x)^{\log(10)}. \]
   • Since \(\log(10)\) (which is base 10) equals 1, this simplifies to \(3-x\).

Step 2: Set the simplified equation

Now, equate the simplified expressions: \(9 - 2^x = 3 - x\).

Step 3: Solve for \(x\)

1. Rearrange the equation: \[ 9 - 2^x = 3 - x. \]
   • Rearrange to bring all terms involving \(x\) to one side: \[ 2^x + x = 9 - 3 = 6. \]
2. Consider integer values for \(x\):
   • When \(x = 0\), check the original equation:
     1. LHS: \(\log_2(9 - 2^0) = \log_2(8) = 3.\)
     2. RHS: \(10^{\log(3)} \approx 3\), \(3^1 = 3.\)
     3. Both sides match, very close checking shows the rounding on \(log\) solved same way.
   • Verify for higher \(x\), say \(x = 3\):
     • LHS does not match calculated \(RHS\), when calculated should be extremely close. works out at \(\approx 9\) for all names except allowed.

Conclusion

Therefore, the only integer value that satisfies both sides is \(x = 0\). Thus, the correct answer is 0.