You may move the terms in x to the left side and then you could use the property of quotient of logarithms having same bases.

log (2x)-log(x-6)=log 5

Use quotient property: log u-log v=log (u/v)

log (2x)-log(x-6)=log [(2x)/(x-6)]

log [(2x)/(x-6)]=log5

The bases of logarithms are the same and the logarithmic function is injective, therefore...

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You may move the terms in x to the left side and then you could use the property of quotient of logarithms having same bases.

log (2x)-log(x-6)=log 5

Use quotient property: log u-log v=log (u/v)

log (2x)-log(x-6)=log [(2x)/(x-6)]

log [(2x)/(x-6)]=log5

The bases of logarithms are the same and the logarithmic function is injective, therefore (2x)/(x-6)=5

Subtract 5 both sides=>

(2x)/(x-6)-5=0

(2x-5x+30)/(x-6)=0

Add terms from numerator:

-3x+30=0

-3x=-30=>x=10

The solution is approved because the only value that is not allowed for x is x=6 because the denominator of the fraction (2x-5x+30)/(x-6) must not be zero.

**Answer: x=10**

We have to solve: log(2x) = log 5 + log(x – 6)

log(2x) = log 5 + log(x – 6)

use the property log a + log b = log (a*b)

=> log(2x) = log (5*(x – 6))

Now we can equate 2x and 5(x - 6)

2x = 5(x - 6)

=> 2x = 5x - 30

=> 3x = 30

=> x = 30/3

=> x = 10

**The solution of the equation is x = 10**