Logarithms questions and answers

Recent questions in Logarithms
Caelan 2021-11-07 Answered

Explain how you would solve the following equation:
\(\displaystyle{\ln{{\left({x}\right)}}}+{\ln{{\left({x}−{5}\right)}}}={\ln{{\left({21}−{x}\right)}}}\)
Describe why you may only choose some of the possible roots of any polynomial you reduce the problem to
Step 1
By using the property of logarithms, simplify the LHS .
\(\displaystyle{\ln{{\left({a}\right)}}}+{\ln{{\left({b}\right)}}}={\ln{{\left({a}{b}\right)}}}\)
\(\displaystyle\text{Simplify the LHS}\)
\(\displaystyle{\ln{{\left({x}\right)}}}+{\ln{{\left({x}-{5}\right)}}}={\ln{{\left({x}{\left({x}-{5}\right)}\right)}}}\)
\(\displaystyle={\ln{{\left({x}^{{2}}-{5}{x}\right)}}}\)
Step 2
Since there is log term on both the sides of the equation, we can take antilog on both sides, simplify and solve the quadratic for x
\(\displaystyle{\ln{{\left({x}^{{2}}-{5}{x}\right)}}}={\ln{{\left({21}-{x}\right)}}}\)
\(\displaystyle{e}^{{{\ln{{\left({x}^{{2}}-{5}{x}\right)}}}}}={e}^{{{\ln{{\left({21}-{x}\right)}}}}}\)
\(\displaystyle{x}^{{2}}-{5}{x}={x}-{21}\)
\(\displaystyle{x}^{{2}}-{5}{x}-{x}-{21}={0}\)
\(\displaystyle{x}^{{2}}-{4}{x}-{21}={0}\)
\(\displaystyle{x}^{{2}}-{7}{x}+{3}{x}-{21}={0}\)
\(\displaystyle{x}{\left({x}-{7}\right)}+{3}{\left({x}-{7}\right)}={0}\)
\(\displaystyle{\left({x}-{7}\right)}{\left({x}+{3}\right)}={0}\)
\(\displaystyle\Rightarrow{x}={7},{x}=-{3}\)
Step 3
There are two solutions possible for value of x, one positive the other negative. However notice that in the given question there is a term ln(x). But since log of a number is defined only for x>0, we discard x=-3.
Hence the answer is x=7.

sjeikdom0 2021-11-06 Answered

Solve \(\displaystyle{{\log}_{{{10}}}{210}}\)

FobelloE 2021-11-05 Answered

243\cdot7^{x-1}=343\cdot3^{x+1}\\

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