Question

4(3^2x+1)+17(3^x)=7

Equations
$$\displaystyle{4}{\left({3}^{{2}}{x}+{1}\right)}+{17}{\left({3}^{{x}}\right)}={7}$$

2021-06-04

We are given: $$\displaystyle{4}{\left({3}^{{2}}{x}+{1}\right)}+{17}{\left({3}^{{x}}\right)}={7}$$
Rewrite $$3^2x+1$$ as $$3^2x\times 3$$ so that: $$\displaystyle{4}{\left({3}^{{2}}{x}{\left({3}\right)}\right)}+{17}{\left({3}^{{x}}\right)}={7}$$
$$\displaystyle{12}{\left({3}^{{2}}{x}\right)}+{17}{\left({3}^{{x}}\right)}={7}$$
Let $$\displaystyle{u}={3}^{{x}}$$ so that
$$\displaystyle{12}{u}^{{2}}+{17}{u}={7}$$
$$\displaystyle{12}{u}^{{2}}+{17}{u}-{7}={0}$$
Factor the left side: $$(3u-1)(4u+7)=0$$
By zero-product property, $$\displaystyle{u}=\frac{{1}}{{3}},-{\left(\frac{{7}}{{4}}\right)}$$
Solve for $$\displaystyle{3}^{{x}}={u}$$:

$$\displaystyle{3}^{{x}}=\frac{{1}}{{3}}$$

$$\displaystyle{3}^{{x}}=-{\left(\frac{{7}}{{4}}\right)}$$
$$\displaystyle{3}^{{x}}={3}^{{-{{1}}}}$$ no solution as $$\displaystyle{3}^{{x}}{>}{0}$$
$$x=-1$$
So, the solution of the equation is: $$x=-1$$