Question

$$\log(x+1)-4\ \log\ x+\log\ 1=0$$

Answer

x=1.220743560791

Solution


Use Rule of One: \(\log{1}=0\).
\[\log{(x+1)}-4\log{x}+0=0\]
Use Power Rule: \(\log_{b}{{x}^{c}}=c\log_{b}{x}\).
\[\log{(x+1)}-\log{{x}^{4}}+0=0\]
Simplify  \(\log{(x+1)}-\log{{x}^{4}}+0\)  to  \(\log{(x+1)}-\log{{x}^{4}}\).
\[\log{(x+1)}-\log{{x}^{4}}=0\]
Use Quotient Rule: \(\log_{b}{\frac{x}{y}}=\log_{b}{x}-\log_{b}{y}\).
\[\log{(\frac{x+1}{{x}^{4}})}=0\]
Simplify  \(\frac{x+1}{{x}^{4}}\)  to  \(\frac{1}{{x}^{3}}+\frac{1}{{x}^{4}}\).
\[\log{(\frac{1}{{x}^{3}}+\frac{1}{{x}^{4}})}=0\]
Use Definition of Common Logarithm: \({b}^{a}=x\) if and only if \(log_b(x)=a\).
\[\frac{1}{{x}^{3}}+\frac{1}{{x}^{4}}=1\]
Multiply both sides by the Least Common Denominator: \({x}^{4}\).
\[x+1={x}^{4}\]
Move all terms to one side.
\[x+1-{x}^{4}=0\]
No root was found algebraically. However, the following root(s) were found by numerical methods.
\[x=-0.724492,1.220744\]
Check solution
When \(x=-0.724492\), the original equation \(\log{(x+1)}-4\log{x}+\log{1}=0\) does not hold true.We will drop \(x=-0.724492\) from the solution set.
Therefore,
\(x=1.220744\)