If log 630- a - log 1524- b then log 1260-A- a B- a w b-2 a b-1C-D- 2 a b-2 a-1-a b-b-1

The correct option is D 2ab+2a 1/ab+b+1lob_6^30 = log_2^30/log_2^6 = 1+log_2^3+log_2^5/1+log_2^3 = a nbsp; nbsp; nbsp; nbsp;...(1) log_15^24 = 3+log ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Linear Combination of Vectors

If log 630= a...

Question

If $l o g_{6}^{30} = a, l o g_{15}^{24} = b t h e n l o g_{12}^{60} =$

\frac{2 a b}{a b + b + 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{a b + b - 1}{2 a b + b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 a (b + 1)}{a b + b}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 a b + 2 a - 1}{a b + b + 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{2 a b + 2 a - 1}{a b + b + 1}$
$l o b_{6}^{30} = \frac{l o g_{2}^{30}}{l o g_{2}^{6}} = \frac{1 + l o g_{2}^{3} + l o g_{2}^{5}}{1 + l o g_{2}^{3}} = a$ ...(1)
$l o g_{15}^{24} = \frac{3 + l o g_{2}^{3}}{l o g_{2}^{3} + l o g_{2}^{5}} = b$ ...(2)
Let $l o g_{2}^{3} = p, l o g_{2}^{5} = q$
From ..(1) 1 + p + q = a(1 + p) ...(3)
From ...(2) 3 + p = b(p + q) $\Rightarrow$ 3 + p = b (a(1+p)-1)
3 + p = ab + abp - b
$p = \frac{a b - b - 3}{1 - a b}$
q = (1+p)(a-1) (From (3))
$q = (a - 1) (1 + \frac{a b - b - 3}{1 - a b})$
$q = \frac{(a - 1) (1 - b - 3)}{(1 - a b)} = \frac{(a - 1) (- b - 2)}{1 - a b}$
$l o g_{12}^{60} = \frac{2 + l o g_{2}^{3} + l o g_{2}^{5}}{2 + l o g_{2}^{3}}$
$= \frac{2 + p + q}{2 + p} = \frac{2 + a (1 + p) - 1}{2 + p}$
$= \frac{1 + a (1 + p)}{2 + p}$
$l o g_{12}^{60} = \frac{1 + a (p + 1)}{p + 2}$
Putting value of P
$l o g_{12}^{60} = \frac{2 a b + 2 a - 1}{a b + b - 1}$

Suggest Corrections

Similar questions

Q. If

{log}_{4} 5 = a

and

{log}_{5} 6 = b

, then

{log}_{3} 2

equal to

Q. Find

n

if:

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{n}

Q. The factors of 8a³ + b³ − 6ab + 1 are

(a) (2a + b − 1) (4a² + b² + 1 − 3ab − 2a)

(b) (2a − b + 1) (4a² + b² − 4ab + 1 − 2a + b)

(c) (2a + b + 1) (4a² + b² + 1 −2ab − b − 2a)

(d) (2a − 1 + b) (4a² + 1 − 4a − b − 2ab)

Q. Solve:

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (1 + a^{2} + b^{2})^{3}

Q. Factorise

8 a^{3} - b^{3} - 4 a x + 2 b x

Join BYJU'S Learning Program

If log306=a,log2415=b then log6012=

If $l o g_{6}^{30} = a, l o g_{15}^{24} = b t h e n l o g_{12}^{60} =$