log2 x=log4 (x+6)?

Question

how to do these sums?

Answer 1

log[base 2](x) = log[base 4](x + 6)

Use the change of base formula on the right hand side; use base 2.

log[base 2](x) = log[base 2](x + 6) / log[base 2](4)

log[base 2](4) = 2, so we have

log[base 2](x) = log[base 2](x + 6) / 2

Multiply both sides by 2.

2 log[base 2](x) = log[base 2](x + 6)

Now, move the 2 inside as a power.

log[base 2](x^2) = log[base 2](x + 6)

Take the antilog of both sides, to eliminate log[base 2].

x^2 = x + 6

Solve as normal.

x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = {3, -2}

However, -2 is an extraneous answer (plugging it into the original equation means taking the log of a negative number, which is strictly not allowed). Therefore, x = 3.

Answer 2

Since log2 x = log4 x^2, then

x^2 = x+6
x^2-x-6=0

x=3 (x cannot = -2 since x>0)

Answer 3

my answer is -12
i have taken the question as log(2x) = log(4(x+6))
then u can equate 2x = 4x + 24
solve and u get x = -12.

if the question is x*(log2) = (x+6)*log4 then
take log 4 = 2 log 2
and u can solve it
u get it as x = 2(x+6)
x = -12

if the question is log x to the base 2 = log (x+6) to the base 4
then the answer is 3,-2
we can't have a negative here hence the answer is 3