solve the equation log(base16)(3x-1)= log(base4)(3x)+log(base4)0.5?

Anish Buchanan 2021-03-07 Answered

solve the equation $$\log(base16)(3x-1)= \log(base4)(3x)+\log(base4)0.5$$?

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Plainmath recommends

• Ask your own question for free.
• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

Expert Answer

unessodopunsep
Answered 2021-03-08 Author has 3701 answers

l$$\displaystyle{o}{g}{\left[{4}\right]}{\left({16}\right)}\cdot{\log{{\left[{16}\right]}}}{4}={1}$$ where [] denotes base, is an example of a fundamental rule in logarithms: $$\displaystyle{\log{{\left[{a}\right]}}}{\left({b}\right)}\cdot{\log{{\left[{b}\right]}}}{\left({a}\right)}={1}$$, so, since $$\displaystyle{\log{{\left[{4}\right]}}}{\left({16}\right)}={2}{\log{{\left[{4}\right]}}}{4}={2},{\log{{\left[{16}\right]}}}{4}=\frac{{1}}{{2}}$$, or, alternatively, $$log[16]4=log[16]\frac{16^{1}}{2}=\frac{1}{2}log[16]16=\frac{1}{2}.$$
Also, $$\displaystyle{\log{{\left[{a}\right]}}}{\left({x}\right)}={\log{{\left[{b}\right]}}}\frac{{{x}}}{{\log{{\left[{b}\right]}}}}$$
(a). $$\displaystyle{\log{{\left[{16}\right]}}}{\left({3}{x}-{1}\right)}={\log{{\left[{4}\right]}}}{\left({3}{x}\right)}+{\log{{\left[{4}\right]}}}{\left({0.5}\right)}⇒{2}{\log{{\left[{4}\right]}}}{\left({3}{x}-{1}\right)}={\log{{\left[{4}\right]}}}{\left({3}{x}\cdot{0.5}\right)}$$. We can equate the logs: $$\displaystyle{\left({3}{x}-{1}\right)}^{{2}}={3}\frac{{x}}{{2}}\rightarrow{9}{x}^{{2}}-{6}{x}+{1}={3}\frac{{x}}{{2}}⇒{18}{x}^{{2}}-{12}{x}-{3}{x}+{2}={0}$$
$$\displaystyle\rightarrow{18}{x}^{{2}}-{15}{x}+{2}={0}\rightarrow{\left({3}{x}-{2}\right)}{\left({6}{x}-{1}\right)}={0}$$
From this $$\displaystyle{x}=\frac{{2}}{{3}}{\quad\text{or}\quad}\frac{{1}}{{6}}.$$ Substitute these values in the original equation: $$\displaystyle{0}=\frac{{1}}{{2}}-\frac{{1}}{{2}}$$. The value $$\displaystyle\frac{{1}}{{6}}$$ cannot be used because it would require the log of a negative number so the only solution is $$Lx=\frac{2}{3}$$

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Plainmath recommends

• Ask your own question for free.
• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.
...