Chapter 5.6, Problem 51AYU

In Problems 41-68, solve each exponential equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places.

$3^{2x}+3^{x+1}–4=0$

To determine

To find: Solve each exponential equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places.

$3^{2x}+{\text{ 3}}^{x\text{ + 1}}–\text{ 4 }=0$

Answer to Problem 51AYU

$x=\text{ 0}$

Explanation of Solution

Given:

$3^{2x}+{\text{ 3}}^{x\text{ + 1}}–\text{ 4 }=0$

Calculation:

Apply (1) to the given equation we have,

$3^{x+1}=3^x\cdot 3$

$3^{2x}+{\text{ 3}}^x\cdot \text{ 3 }-\text{ 4 }=0$

Apply (2) to the above equation we get,

${\left(3^x\right)}^2+3^x\cdot 3-4=0$

Rewrite the equation as $3^x=u$.

$u^2+u\cdot 3-4=0$

Factorizing the above equation we get,

$u=\text{ 1}$ and $u=-4$.

Since $u={\text{ 3}}^x,{\text{ 3}}^x=\text{ 1}$.

Using the condition that if $f\left(x\right)=g\left(x\right)$ then $\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$ in above equation,

$\ln (3^x)=\ln (1)$

Applying (3) to the above equation we get,

$x\ln \left(3\right)=\ln \left(1\right)$

$x\ln \left(3\right)=\text{ 0}$   [since $\log \left(1\right)=0$]

$x=\text{ 0}$

$3^x=-4$, No solution for $x\in ℝ$.