Question

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

Equations
ANSWERED
asked 2021-06-03
\(\displaystyle{4}{\left({3}^{{2}}{x}+{1}\right)}+{17}{\left({3}^{{x}}\right)}={7}\)

Answers (1)

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: \(x=-1\)

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