Verified answer

Take the number in front of the logarithm and place it back in its argument as a power:

log[2](x) = log[4](5) + log[2](27)

Subtract log[2](27) from both sides:

log[2](x) - log[2](27) = log[4](5)

When subtracting logarithms with common bases, divide their arguments from each other:

log[2](x/27) = log[4](5)

The bases aren't common, so apply the change of base formula:

log[2](x/27) = log[2](5)/log[2](4)

Log[2](4) is 2 because 2^2 is 4 so change that:

log[2](x/27) = log[2](5)/2

Multiply both sides by 2 to cancel off the denominator:

2log[2](x/27) = log[2](5)

Using the same property as before:

log[2](x^2/729) = log[2](5)

Now that the bases are common, you can drop the logs:

(x^2)/729 = 5

Square root both sides:

x/27 = √5

Multiply both sides by 27 to cancel off the denominator:

x = 27√5

log₂(x) = log₄(5)+ 3log₂(3)

log₄(5) = log₂(5)/log₂(4) = log₂(5)/log₂(2²) = log₂(5)/2

So, the given equation can be written as:

log₂(x) = log₂(5)/2 + 3log₂(3)

log₂(x) = log₂(5)^(1/2) + log₂(3³)

log₂(x) = log₂(√5) + log₂(27)

log₂(x) = log₂(27√5)

x = 27√5

log2(x)=log4(5)+3log2(3)

log2(x)-3log2(3)=log4(5)

log2(x)-log2(27)=log4(5)

log2(x/27)=log4(5)

2^(log4(5))=x/27

sqrt(4^(log4(5)))=x/27

sqrt(5)=x/27

x=27 sqrt(5)

Log[2](x) = Log[4](5) + 3.Log[2](3) → you know that: Log[a](x) = Ln(x)/Ln(a)

Ln(x)/Ln(2) = [Ln(5)/Ln(4)] + [3.Ln(3)/Ln(2)] → you know that: Ln(x^a) = a.Ln(x)

Ln(x)/Ln(2) = [Ln(5)/Ln(2²)] + [Ln(3³)/Ln(2)]

Ln(x)/Ln(2) = [Ln(5)/2.Ln(2)] + [Ln(27)/Ln(2)]

Ln(x) = [Ln(5)/2] + Ln(27)

Ln(x) = (1/2).Ln(5) + Ln(27)

Ln(x) = Ln(√5) + Ln(27) → you know that: Ln(a) + Ln(b) = Ln(ab)

Ln(x) = Ln(√5 * 27)

→ x = 27√5