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1 year ago

If log (m + n) = log m+log n, show that m =n/n-1

Mathematics

1 answer:

Sedbober [7]1 year ago

3 0

log (m + n) = log m+ log n and proved it m =n/n-1

Given;

If log (m + n) = log m+ log n

To show that the m =n/n-1

Now, According to the question:

We know that,

Log (m + n) = log m + log n

Log (m + n ) = log (mn). [log a + log b = log ab ]

Cancelling the log on both sides.

then,

m + n = mn

=> n = mn - m

=> n = m (n - 1)

=> m = n / n - 1

Hence Proved

log (m + n) = log m+ log n and proved it m =n/n-1

What is Logarithm?

A logarithm is the power to which a number must be raised in order to get some other number (see Section 3 of this Math Review for more about exponents). For example, the base ten logarithm of 100 is 2, because ten raised to the power of two is 100: log 100 = 2.

Learn more about Logarithm at:

brainly.com/question/16845433

#SPJ1

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If 13cos theta -5=0 find sin theta +cos theta / sin theta -cos theta

Ivahew [28]

Step-by-step explanation:

We have to find the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0.

$\red{\frak{Given}} \begin{cases} & \sf {13\ cos \theta\ -\ 5\ =\ 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big\lgroup Can\ also\ be\ written\ as \big\rgroup} \\ & \sf {cos \theta\ =\ {\footnotesize{\dfrac{5}{13}}}} \end{cases}$

Here, we're asked to find out the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0. In order to find the solution we're gonna use trigonometric ratios to find the value of sinθ and cosθ. Let us consider, a right angled triangle, say PQR.

Where,

PQ = Opposite side
QR = Adjacent side
RP = Hypotenuse
∠Q = 90°
∠C = θ

As we know that, 13 cosθ - 5 = 0 which is stated in the question. So, it can also be written as cosθ = 5/13. As per the cosine ratio, we know that,

$\rightarrow {\underline{\boxed{\red{\sf{cos \theta\ =\ \dfrac{Adjacent\ side}{Hypotenuse}}}}}}$

Since, we know that,

cosθ = 5/13
QR (Adjacent side) = 5
RP (Hypotenuse) = 13

So, we will find the PQ (Opposite side) in order to estimate the value of sinθ. So, by using the Pythagoras Theorem, we will find the PQ.

Therefore,

$\red \bigstar {\underline{\underline{\pmb{\sf{According\ to\ Question:-}}}}}$

$\rule{200}{3}$

$\sf \dashrightarrow {(PQ)^2\ +\ (QR)^2\ =\ (RP)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ (5)^2\ =\ (13)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ 25\ =\ 169} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 169\ -\ 25} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 144} \\ \\ \\ \sf \dashrightarrow {PQ\ =\ \sqrt{144}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{PQ\ (Opposite\ side)\ =\ 12}}}}_{\sf \blue{\tiny{Required\ value}}}}$

∴ Hence, the value of PQ (Opposite side) is 12. Now, in order to determine it's value, we will use the sine ratio.

$\rightarrow {\underline{\boxed{\red{\sf{sin \theta\ =\ \dfrac{Opposite\ side}{Hypotenuse}}}}}}$

Where,

Opposite side = 12
Hypotenuse = 13

Therefore,

$\sf \rightarrow {sin \theta\ =\ \dfrac{12}{13}}$

Now, we have the values of sinθ and cosθ, that are 12/13 and 5/13 respectively. Now, finally we will find out the value of the following.

$\rightarrow {\underline{\boxed{\red{\sf{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}}}}}}$

By substituting the values, we get,

$\rule{200}{3}$

$\sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\Big( \dfrac{12}{13}\ +\ \dfrac{5}{13} \Big)}{\Big( \dfrac{12}{13}\ -\ \dfrac{5}{13} \Big)}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{13} \times \dfrac{13}{7}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{\cancel{13}} \times \dfrac{\cancel{13}}{7}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{7}}}}}_{\sf \blue{\tiny{Required\ value}}}}$