Number - Exponential equations without logarithms

Solve for $x$: $2^{x+3} = 64$

$2^{x+3} = 2^6$ $x + 3 = 6$ $x = 3$

First, express $64$ as a power of $2$, which is $2^6$. Since the bases are equal, the exponents must be equal.

Solve for $x$: $9^{x-1} = \left(\frac{1}{27}\right)^{x}$

$(3^2)^{x-1} = (3^{-3})^x$ $3^{2x-2} = 3^{-3x}$ $2x - 2 = -3x$ $5x = 2$ $x = \frac{2}{5}$

Identify a common base for $9$ and $27$, which is $3$. Rewrite $9$ as $3^2$ and $\frac{1}{27}$ as $3^{-3}$. Apply the power of a power rule and solve for $x$.

Solve for $x$: $4^{x} \times 8^{x-1} = 2$

$(2^2)^x \times (2^3)^{x-1} = 2^1$ $2^{2x} \times 2^{3x-3} = 2^1$ $2^{2x + 3x - 3} = 2^1$ $5x - 3 = 1$ $5x = 4$ $x = \frac{4}{5}$

Convert all terms to base $2$. Use the multiplication law of indices $a^m \times a^n = a^{m+n}$ to combine the terms on the left side, then equate the exponents.