Functions - Exponential equations

Solve for $x$: $3^{2x-1} = 27$

$3^{2x-1} = 3^3$ $2x - 1 = 3$ $2x = 4$ $x = 2$

Since $27$ can be written as $3^3$, we can equate the exponents because the bases are now the same.

Solve for $x$: $5^x = 12$, giving your answer to 3 significant figures.

$\ln(5^x) = \ln(12)$ $x \ln 5 = \ln 12$ $x = \frac{\ln 12}{\ln 5}$ $x \approx 1.54$

Since the bases cannot be made the same easily, we take the natural logarithm of both sides and use the power rule to isolate $x$.

Solve the equation $4^x - 6(2^x) + 8 = 0$.

Let $u = 2^x$. Then $4^x = (2^2)^x = (2^x)^2 = u^2$. $u^2 - 6u + 8 = 0$ $(u - 4)(u - 2) = 0$ $u = 4 \text{ or } u = 2$ Substituting back $u = 2^x$: $2^x = 4 \implies x = 2$ $2^x = 2 \implies x = 1$

This is an equation in quadratic form. By substituting $u = 2^x$, we transform it into a standard quadratic equation, solve for $u$, and then solve for $x$.

1. Rewrite the equation using a common base: $(3^2)^x - 12(3^x) + 27 = 0$.
2. Use the power rule: $(3^x)^2 - 12(3^x) + 27 = 0$.
3. Let $u = 3^x$. The equation becomes $u^2 - 12u + 27 = 0$.
4. Factor the quadratic: $(u - 3)(u - 9) = 0$.
5. Solve for $u$: $u = 3$ or $u = 9$.
6. Substitute back: $3^x = 3$ or $3^x = 9$.
7. Solving these gives $x = 1$ or $x = 2$.

This is a quadratic-type exponential equation. By substituting $u = 3^x$, we transform a transcendental equation into a manageable algebraic one.

1. Take the natural logarithm of both sides: $\ln(2^{x+1}) = \ln(7^x)$.
2. Use the power property: $(x+1)\ln 2 = x \ln 7$.
3. Expand the left side: $x \ln 2 + \ln 2 = x \ln 7$.
4. Rearrange to isolate $x$: $\ln 2 = x \ln 7 - x \ln 2$.
5. Factor out $x$: $\ln 2 = x(\ln 7 - \ln 2)$.
6. Solve for $x$: $x = \frac{\ln 2}{\ln 7 - \ln 2}$.
7. Calculate: $x \approx 0.553$.

Since the bases 2 and 7 cannot be written as powers of a common integer base, we use logarithms to isolate the variable in the exponent.