Quantitative Ability Tech Test Logarithms Test 1

$Log_{b}b$ = 1. $Hence \:log_{5}5$ = 1

$Log_{b}1$ = 0. $Hence\: log_{5}1$ = 0

log(a × b) = log a + log b

similarly, log(a × b × c) = log a + log b + log c

Hence log(2×4×6) = log 2 + log 4 + log 6

log(3+4) = log(3 × 4) is wrong

LHS = log(3+4) = log 7

RHS = log(3 × 4) = log(12)

log 7 ≠ log 12

$log_{5}1024$

$\dfrac{log\:1024}{log\:5}$

=$\dfrac{log\left(2^{10}\right)}{log\left(\dfrac{10}{2}\right)}$

=$\dfrac{10\:log\left(2\right)}{log\:10-log\:2}$

=$\dfrac{10\times0.3010}{1-0.3010}$

=$\dfrac{3.01}{0.699}$

=$\dfrac{3010}{699}$

=4.31

$log(648)^{5}$

= 5 log(648)

= 5 log(81 × 8)

= 5[log(81) + log(8)]

=$5 [log(3^{4}) + log(2^{3})] $

=5[4log(3) + 3log(2)]

= 5[4 × 0.4771 + 3 × 0.30103]

= 5(1.9084 + 0.90309)

= 5 × 2.81149

≈ 14.05

ie, $log(648)^{5}$ ≈ 14.05

ie, its characteristic = 14

Hence, number of digits in $(648)^{5}$ = 14+1 = 15

$log_{4}x+log_{2}x$=12

=>$\dfrac{log\:x}{log\:4}+\dfrac{log\:x}{log\:2}$=12

=>$\dfrac{log\:x}{log\:2^2}+\dfrac{log\:x}{log\:2}$=12

=>$\dfrac{log\:x}{2\:log\:2}+\dfrac{log\:x}{log\:2}$=12

=>$\dfrac{log\:x+2\:log\:x}{2\:log\:2}$=12

=>$\dfrac{3\:log\:x}{2\:log\:2}$=12

Therefore,

logx=$\dfrac{12\times2\:log\:2}{3}$

=8 log 2

=$log(2^8)$

=log(256)

x=256

Let $log_{(.001)} (100)$ =p

$(.001)^p$=100

$\left(\dfrac{1}{1000}\right)^p$=100

$\left(\dfrac{1}{10^3}\right)^p$=$10^2$

$[\left(10\right)^{-3}]^p$=$10^2$

$\left(10\right)^{-3p}$=$10^2$

-3p=2

p=$\dfrac{-2}{3}$

$log_{3} 4\times log_{4} 5 \times log_{5} 6\times log_{6} 7 \times log_{7} 8 \times log_{8} 9 \times log_{9} 9$

$\dfrac{log \:4}{log\:3} \times \dfrac{log \:5}{log\:4} \times \dfrac{log \:6}{log\:5} \times \dfrac{log \:7}{log\:6} \times \dfrac{log \:8}{log\:7} \times \dfrac{log \:9}{log\:8} \times 1$

=$\dfrac{log \:9}{log\:3}$

=$\dfrac{log \:3^2}{log\:3}$

=$\dfrac{2log \:3}{log\:3}$

=2

$log_{12} 27$ = a

=> log 27/ log 12 = a

=> $log 3^{3}$ / $log (3 * 2^{2})$ =a

=> 3 log 3 / log 3 + 2 log 2 = a

=> (log 3 + 2 log 2)/ 3 log 3 = 1/a

=> (log 3/ 3 log 3) + (2 log 2/ 3 log 3) = 1/3

=> (2 log 2)/ (3 log 3) = 1/a – 1/3 = (3-a)/ 3a

=> log 2/ log 3= (3-a)/3a

=> log 3 = (2a/3-a)log2

$log_{16} 16 $= log 16/ log 6

= $log 2^{4}/ log (2*3) $

= 4 log 2/ (log 2 + log 3)

= 4(3-a)/ (3+a)

$log_{a} (ab)$ = x

=> log b/ log a = x

=> (log a + log b)/ log a = x

1+ (log b/ log a) = x

=> log b/ log a = x-1

log a/ log b = 1/ (x-1)

=> 1+ (log a/ log b) = 1 + 1/ (x-1)

(log b/ log b) + (log a/ log b) = x/ (x-1) => (log b + log a)/ log b = x/ (x-1)

=>log (ab)/ log b = x/(x-1) => $log_{b} (ab)$ = x/(x-1)

= $\dfrac{1}{3}log_{10}125−2log_{10}4+log_{10}32$

=$log_{10}(125^{1/3})−log_{10}(4^2)+log_{10}32$

=$log_{10}5−log_{10}16+log_{10}32$

=$log_{10}\left(\dfrac{5\times32}{16}\right)$

=$log_{10}10$

=1

If base is not mentioned, then always remember to take it as 10.

Hence, in the given expression, assume base as 10

We are given, $[log_{10} 2 + log_{10} (4x + 1)$ = $log_{10} (x + 2) + 1]$

$[log_{10} 2 + log_{10} (4x + 1)$ = $log_{10} (x + 2) + 1]$

$[log_{10} 2 + log_{10} (4x + 1)$ = $log_{10} (x + 2) + log_{10}10]$

Now, Use the product rule: $log_{a}(xy)$ = $log_{a}x + log_{a}y$

$[log_{10} 2 (4x + 1)]$ = $[log_{10} 10(x + 2)]$

(4x + 1) = (5x + 10)

4x + 1 = 5x + 10

x = - 9