Question

Express 1/2log(x+y)--1/2log(x-y) as single logarithm​

Answer 1

Is 1/2logx+1/2logy equivalent to 1/2log(x+y)?

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2 ANSWERS



Nicholas Cooper, Mathematics is like art: the more abstract it is, the less some see the beauty

Answered Dec 6, 2017

So, you are joining logarithms together:

12logx+12logy12log⁡x+12log⁡y

The rule of exponents is that:

xa+b=(xa)(xb)xa+b=(xa)(xb)

So, logarithms get the “opposite rule”:

log(xy)=logx+logylog⁡(xy)=log⁡x+log⁡y

So we know that the expression you gave us is equivalent to this:

12(logx+logy)12(log⁡x+log⁡y)

Which means that on combining the logarithms into a single logarithm:

12log(xy)12log⁡(xy)

Furthermore, one more identity that you may be aware of, is that this is also equivalent to:

log((xy)12)=logxy−−√log⁡((xy)12)=log⁡xy

But only if x,y∈Rx,y∈R. If x and y can be complex, then there are an infinite number of solutions, and the square root should be avoided (it’s still correct, but may cause you to miss some of the infinite results that may be available).

Answer 2

The required expression is $\log (\dfrac{\sqrt{x+y}}{\sqrt{x-y}})$.

Step-by-step explanation:

The given expression is

$\dfrac{1}{2}\log (x+y)-\dfrac{1}{2}\log (x-y)$

We need to expression the given problem as single logarithm​.

Using properties of logarithm we get

$\log (x+y)^{\frac{1}{2}}-\log (x-y)^{\frac{1}{2}}$               $[\because \log a^b=b\log a]$

$\log \sqrt{(x+y)}-\log \sqrt{(x-y)}$

$\log (\dfrac{\sqrt{x+y}}{\sqrt{x-y}})$            $[\because \log (\dfrac{a}{b})=\log a-\log b]$

Therefore, the required expression is $\log (\dfrac{\sqrt{x+y}}{\sqrt{x-y}})$.

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