Question

Find the value of

& i){{\log }_{2}}32 \\\ & ii){{\log }_{5}}3\sqrt{5} \\\ & iii){{\log }_{\sqrt{2}}}64 \\\ & iv){{\log }_{10}}100000 \\\ \end{aligned}$$

Answer 1

The above question uses the some elementary concepts of logarithm.
${{\log }_{a}}N=x$ read as log of N to the base a if a = e, we write $\ln N\Rightarrow {{\log }_{e}}N$ (Natural log).
The three major properties of logarithm is used in this question:

& \to {{\log }_{a}}a=1 \\\ & \to {{\log }_{a}}{{x}^{p}}=p{{\log }_{a}}x;x\text{ }>\text{ }0 \\\ & \to {{\log }_{{{a}^{q}}}}x=\dfrac{1}{q}{{\log }_{a}}x;x\text{ }>\text{ }0 \\\ \end{aligned}$$ _**Complete step-by-step solution:**_ Now, we have given data as: $$\left( \text{i} \right){{\log }_{2}}32$$ Here, we can write $32={{2}^{5}}$ therefore, we get ${{\log }_{2}}{{2}^{5}}$ Now, according to property ${{\log }_{a}}{{x}^{p}}=p{{\log }_{a}}x$ here, $${{\log }_{2}}{{2}^{5}}=5{{\log }_{2}}2$$ Now, we know property ${{\log }_{a}}a=1$ hence, $$5{{\log }_{2}}2=5\times 1=5$$ Hence, answer of (i) is $${{\log }_{2}}32=5$$ Now, coming to the second part of the question i.e. $$\left( ii \right){{\log }_{5}}3\sqrt{5}$$ We know that, $$3\sqrt{5}={{\left( 5 \right)}^{\dfrac{1}{3}}}\left[ \because n\sqrt{y}={{\left( y \right)}^{\dfrac{1}{n}}} \right]$$ Therefore, we get ${{\log }_{5}}{{\left( 5 \right)}^{\dfrac{1}{3}}}$ Now, apply the property ${{\log }_{a}}{{x}^{p}}=p{{\log }_{a}}x$ $$\Rightarrow {{\log }_{5}}{{\left( 5 \right)}^{\dfrac{1}{3}}}=\dfrac{1}{3}{{\log }_{5}}5$$ By applying property ${{\log }_{a}}a=1$ we get: $$\Rightarrow \dfrac{1}{3}{{\log }_{5}}5=\dfrac{1}{3}\times 1=\dfrac{1}{3}$$ Hence, answer of (ii) is $${{\log }_{5}}3\sqrt{5}=\dfrac{1}{3}$$ Now, coming to the third part of the question, i.e. $$\left( iii \right){{\log }_{\sqrt{2}}}64$$ Now, this part uses the different properties as compared to the previous two parts. Here, we can write $$\begin{aligned} & \sqrt{2}={{\left( 2 \right)}^{\dfrac{1}{2}}} \\\ & \therefore {{\log }_{{{\left( 2 \right)}^{\dfrac{1}{2}}}}}64 \\\ \end{aligned}$$ Now, using property ${{\log }_{{{a}^{q}}}}x=\dfrac{1}{q}{{\log }_{a}}x$ We have, $${{\log }_{{{\left( 2 \right)}^{\dfrac{1}{2}}}}}64=2{{\log }_{2}}64$$ Now, 64 can be written as ${{2}^{6}}$ $$\therefore 2{{\log }_{2}}{{\left( 2 \right)}^{6}}$$ Now, again using property ${{\log }_{a}}{{x}^{p}}=p{{\log }_{a}}x$ we have $2\times 6{{\log }_{2}}\left( 2 \right)$ Finally, by applying property ${{\log }_{a}}a=1$ we get: $$\Rightarrow 2\times 6\times {{\log }_{2}}\left( 2 \right)=2\times 6\times 1=12$$ Hence, answer of (iii) is $${{\log }_{\sqrt{2}}}64=12$$ Now, coming to the fourth part of the question i.e. $$\left( iv \right){{\log }_{10}}100000$$ Here, we can write 100000 as ${{10}^{5}}$ $$\therefore {{\log }_{10}}{{\left( 10 \right)}^{5}}$$ Using property ${{\log }_{a}}{{x}^{p}}=p{{\log }_{a}}x$ we have: $$\Rightarrow {{\log }_{10}}{{\left( 10 \right)}^{5}}=5{{\log }_{10}}10$$ Finally, using property ${{\log }_{a}}a=1$ we have: $$\Rightarrow 5{{\log }_{10}}10=5\times 1=5$$ Hence, answer of (iv) is $${{\log }_{10}}100000=5$$ **Note:** The most common mistake that may occur in logarithm is "the expression ${{\log }_{a}}{{x}^{p}}$ is written as ${{\left( {{\log }_{a}}x \right)}^{p}}$". We can’t write ${{\log }_{a}}{{x}^{p}}={{\left( {{\log }_{a}}x \right)}^{p}}$. The other important thing is necessary conditions. Like we have ${{\log }_{a}}N=x$ then $N\text{ }>\text{ }0,a\text{ }>\text{ }0,a\ne 1$ Some general points like we have been given logN (where base of log is not given in the question) then we will take base as 'e'. Therefore, ${{\log }_{e}}N=\ln N$