Solving In and write as a single logarithm

Question

In 10 + 3 in 4 - 5 in 2

then

2 log 6 = log 3 + log (x-3)

Answer 1

For the first problem we use the properties of logarithms.

ln 10 + 3 ln 4 - 5 ln 2

whenever there is a "coefficient" in front of a log function it acts as the exponent inside the function.

ln 10 + ln 4³ – ln 2⁵

ln 10 + ln 64 – ln 32

now where there is addition between the log functions it becomes multiplication of the values within

ln (10*64) – ln 32

ln 640 – ln 32

and in a similar fashion, subtraction indicates division

ln (640 / 32)

ln 20

Done.

Follow the same rules to solve for x in the equation.

2 log 6 = log 3 + log (x – 3)

log 6² = log [3(x – 3)]

log 36 = log (3x – 9)

now set the values inside the log function equal to one another

36 = 3x – 9

and now solve with algebra



Answer 2

Adding logarithms with the same base can be rewritten as the logarithm of the product of the arguments.

Ex: lnx + lny = ln(xy)

Ex: lnx - lny = ln(x/y)

The other thing to remember with logarithms is how a power on the argument can be rewritten as a constant out front.

Ex: ln(x²) = 2lnx

With those two rules in mind:

ln(10) + 3ln(4) - 5ln(2) = ln(10) + ln(4³) - ln(2⁵) = ln(10*64/32) = ln(20)

Your second problem:

2log(6) = log(3)+ log(x-3)

log(6²) = log(3(x-3)) Since these two expressions are equal, their arguments must also be equal.

36 = 3(x-3)

12 = x - 3

x = 15

Answer 3

2 log 6 = log 3 + log (x-3)

x =15