Solved Problems on Logarithms

In this section, we will solve problems on logarithms. For the definition of a logarithm, please click on this link.

At first, we will recall the main formulas of logarithms:

Formula 1:  $\log_a(MN)=\log_a M + \log_a N$

Formula 2:  $\log_a(M/N)=\log_a M – \log_a N$

Formula 3:  $\log_a M^n=n \log_a M$

Formula 4:  $\log_a M=\log_b M \times \log_a b$

Problems and Solutions of logarithms:

Problem 1:  Show that $\log_3 \log_2 8=1$
At first, we calculate $\log_2 8$.
Now, $\log_2 8=\log_2 2^3$
$=3 \log_2 2 \quad$ $[\because \log_a b^k=k \log_a b]$
$=3 \cdot 1 \quad$ $[\because \log_a a=1]$ 
$\therefore$ LHS $= \log_3 \log_2 8$
$=\log_3 3$
$=1$ (proved)


Problem 2:  Find the value of  $\log_2 \log_2 \log_2 16$
Using the above Formula 3, we have
 $\log_2 16=\log_2 2^4=4 \log_2 2=4$
So LHS $=\log_2 \log_2 \log_2 16$
$=\log_2 \log_2 4$
$=\log_2 \log_2 2^2$
$=\log_2 2 \log_2 2$
$=1 \cdot 1 \quad$ $[\because \log_a a=1]$
$=1$ (proved)
Problem 3: Express the following logarithms in terms of indices
(i) $\log_x 1=0$
(ii) $\log_A B=C$
(iii) $\log_3 81=4$
(iv) $\log_2 \frac{1}{32}=-5$
We will use the fact: $\log_a b=x \iff a^x=b$
(i) $x^0=1$
(ii) $A^C=B$
(iii) $3^4=81$
(iv) $2^{-5}=\frac{1}{32}$
Problem 4: Find the value of $x$ when $x=\log_2 64$
Given that $x=\log_2 64$
$\Rightarrow 2^x=64$
$\Rightarrow 2^x=2^6$
Equating the exponents, we get that
$x=6$ ans.
Problem 5: Find the base when the logarithm of $343$ is $3$
Let $x$ be the desired base.
According to the question, $\log_x 343=3$
$\Rightarrow x^3=343$
$\Rightarrow x^3=7^3$
As we have the same powers, we obtain that
So the desired base is $7$ ans.
Problem 6: Prove that $\log (1+2+3)=\log 1+\log 2 + \log 3$
LHS $=\log(1+2+3)$
$=\log 6$
$=\log(1 \cdot 2 \cdot 3)$
$=\log 1 +\log 2 + \log 3=$RHS (proved)
$[\because \log(abc)=\log a + \log b + \log c]$
Problem 7:  Show that $a^{\log_a x}=x$
Suppose that $\log_a x=z$
So by the definition of logarithms, we have
$a^z=x \cdots (i)$
Now, LHS $=a^{\log_a x}$
$=x$ [by equation (i)]
$=$ RHS (proved)
Problem 8: Find the base when the logarithm of $1/3$ is $\frac{-1}{3}$
Let $x$ be the base.
Given that $\log_x \frac{1}{3}=\frac{-1}{3}$
$\Rightarrow x^{-1/3}=1/3$
Taking cubes on both sides, we get that
$\Rightarrow  x^{-1/3 \times 3}=\frac{1}{3^3}$
$\Rightarrow  x^{-1}=\frac{1}{27}$
$\Rightarrow  1/x=\frac{1}{27}$
Taking reciprocals, we have
So the base is $27$ ans.
Problem 9: Find the value of $\log_8 27$ when $\log_2 3=a$
Note that $\log_8 27=\log_8 3^3$
$=3 \log_8 3$ $[\because \log_a b^k=k \log_a b]$
$=3 \cdot \frac{1}{\log_3 8}$ $[\because \log_b a=\frac{1}{\log_a b}]$
$=3 \cdot \frac{1}{\log_3 2^3}$
$=3 \cdot \frac{1}{3 \log_3 2}$
$=\frac{1}{\log_3 2}$
$=\log_2 3$
So the desired answer is $a$.
Problem 10: Evaluate $\log_3 5 \times \log_{25} 27$
First we calculate $\log_{25} 27$
Now $\log_{25} 27=\log_{25} 3^3$
$=3 \log_{25} 3$ $[\because \log_a b^k=k \log_a b]$
 $=3 \cdot \frac{1}{\log_3 25}$ $[\because \log_b a=\frac{1}{\log_a b}]$
$=3 \cdot \frac{1}{\log_3 5^2}$
$=3 \cdot \frac{1}{2\log_3 5}$
$=3/2 \cdot \frac{1}{\log_3 5}$
Therefore, $\log_3 5 \times \log_{25} 27$
$=\log_3 5 \times 3/2 \times \frac{1} {\log_3 5}$
$=3/2$ ans.
Problem 11: Prove that $a^{5\log_a x}=x^5$
Note that $5 \log_a x=\log_a x^5$
Now, LHS $=a^{5 \log_a x}$
$=a^{\log_a x^5}$
$=x^5=$ RHS (proved)
[by Problem 7]
Problem 12: Prove that $\log_4 2 \times \log_2 3= \log_4 5 \times \log_5 3$
We will use $\log_a M=\log_b M \times \log_a b$
LHS $=\log_4 2 \times \log_2 3 =\log_4 3$
RHS $=\log_4 5 \times \log_5 3=\log_4 3$
So LHS=RHS  (proved)
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