204/VB02. problem / Math - Solved Math Problems

MR-204 / 1071b. problem

Solve the following exponential equation:

$4^{x + 1} + 4^{x} = 320$

Source: Vene T. Bogoslavov: Collection of Solved Problems #2 - 1071.b.)

Algebra → Exponential equations and inequalities → Exponential equations

4^{x + 1} + 4^{x} = 320

4^{x} \cdot 4 + 4^{x} = 320

4^{x} (4 + 1) = 320

4^{x} \cdot 5 = 320

4^{x} = \frac{320}{5}

4^{x} = 64

Write number 64 as a factor of 4: $64 = 4^{3}$

4^{x} = 4^{3}

x = 3