Question

If $\sf{log_{8} {a}^{3}}$ + $\sf{log_{4} {b}^{2}}$ = 5, then
(A) the last digit of ab + 1 is even
(B) the last digit of ab is odd
(C) ab + 17 is a perfect square
(D) ab + 1 is a perfect cube

Single option correct

[Ans. (C).] Explanation needed.​

Answer 1

Solutìon :

(C) ab + 17 is a perfect square

Explanatìon :

We have ,

$\sf\log_{8}a^3+\log_{4}b^2=5$

Use property:

$\sf\blue{\log(a)^n=n\log(a)}$ then ,

$\sf\implies3\log_{8}a+2\log_{4}b=5$

$\sf\implies3\log_{2^3}a+2\log_{2^2}b=5$

Use property :

$\sf\green{\log_{a{}^{n}}x=\dfrac{1}{n}\times\log_{a}x}$ then ,

$\sf\implies3\times\dfrac{1}{3}\log_2a+2\times\dfrac{1}{2}\log_2b=5$

$\sf\implies\dfrac{3}{3}\log_2a+\dfrac{2}{2}\log_2b=5$

$\sf\implies\log_2a+\log_2b=5$

Use property :

$\sf\pink{\log(a)+\log(b)=\log(ab)}$ then

$\sf\implies\log_2(ab)=5$

$\sf\implies2^5=ab$

$\sf\implies\:ab=32...(1)$

Case : 1

When 1 is added to equation (1)

$\sf\:ab+1=32+1$

$\sf\:ab+1=33$

the last digit of ab + 1 is odd and it's also not a perfect cube , hence a) & d) options are wrong

Case :2

$\sf\:ab+1=32$

the last digit of ab + 1 is even,

Hence , b ) option is wrong

Case :3

Add 17 in Equation (1)

$\sf\:ab+17=32+17$

$\sf\:ab+17=49$

$\sf\:ab+17=7^2$

ab +17 is a perfect square of 7

Hence , option c ) is correct

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Rules of Logarithm :

• Basic rules

$\sf\:1)\log(a) + \log(b) = \log(ab)$

$\sf\:2)\log( \frac{a}{b} ) = \log(a) - \log(b)$

$\sf\:3)\log(a) {}^{n} = nlog(a)$

$\sf\:4)\log_{a}(a) = 1$

$\sf\:5)\log_{a}(1)=0$

$\sf\:6)\log_{a{}^{n}}x=\dfrac{1}{n}\times\log_{a}x$

$\sf\:7)\log_{a{}^{n}}x{}^{m}=\dfrac{m}{n}\times\log_{a}$

•Base changing rule

$\sf\:\log_{a}x = \dfrac{\log_{b}x}{\log_{b}a}$

Answer 2

Answer:

When 1 is added to equation (1)

ab+1=32+1ab+1=32+1

ab+1=33ab+1=33

the last digit of ab + 1 is odd and it's also not a perfect cube , hence a) & d) options are wrong