If log 18 to the base 12=a and log 54 to the base 24=b,prove that ab+5(a-b)=1?

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If log 18 to the base 12=a and log 54 to the base 24=b,prove that ab+5(a-b)=1?

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Answer 1 · 2017-07-11T18:13:06

see explanation

Explanation:

Please check to make sure I've read the problem correctly. Logarithms may be written wrong...

${log}_{12} (18) = a$ $\log_12(18)=a$
${log}_{24} (54) = b$ $\log_24(54)=b$

$a b + 5 (a - b) ? = 1$ $ab+5(a-b)\stackrel{?}{=}1$
$({log}_{12} (18) \cdot {log}_{24} (54)) + 5 ({log}_{12} (18) - {log}_{24} (54)) \sqrt{} = 1$ $(\log_12(18)\cdot\log_24(54))+5(\log_12(18)-\log_24(54))\stackrel{\color(olive)(\sqrt{})}{=}1$

Answer 2 · 2017-07-12T09:15:30

Please see below.

Explanation:

As ${log}_{12} (18) = a$ $log_12(18)=a$ , we have $12^{a} = 18$ $12^a=18$

i.e. ${(2^{2} \times 3)}^{a} = 2 \times 3^{2}$ $(2^2xx3)^a=2xx3^2$

or $2^{2 a - 1} \times 3^{a - 2} = 1$ $2^(2a-1)xx3^(a-2)=1$

similarly as ${log}_{24} (54) = b$ $log_24(54)=b$ , we have $24^{b} = 54$ $24^b=54$

i.e. ${(2^{3} \times 3)}^{b} = 2 \times 3^{3}$ $(2^3xx3)^b=2xx3^3$

or $2^{3 b - 1} \times 3^{b - 3} = 1$ $2^(3b-1)xx3^(b-3)=1$

Comparing the two $2 a - 1 = 3 b - 1$ $2a-1=3b-1$ or $2 a - 3 b = 0$ $2a-3b=0$ ........(A)

and $a - 2 = b - 3$ $a-2=b-3$ or $a - b + 1 = 0$ $a-b+1=0$ i.e. $2 a - 2 b + 2 = 0$ $2a-2b+2=0$ ........(B)

Subtracting (A) from (B), we get

$b + 2 = 0$ $b+2=0$ i.e. $b = - 2$ $b=-2$

and $a = b - 1 = - 3$ $a=b-1=-3$

hence $a b + 5 (a - b)$ $ab+5(a-b)$

$= (- 3) \times (- 2) + 5 \times (- 1) = 6 - 5 = 1$ $=(-3)xx(-2)+5xx(-1)=6-5=1$

(Note that (B) gives $a - b = - 1$ $a-b=-1$ .)

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If log 18 to the base 12=a and log 54 to the base 24=b,prove that ab+5(a-b)=1?

2 Answers

Explanation:

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