Question

How do you solve `log_2 x = log_4 (x+6)`?

Answer 1

Using the change of base rule we have that

#logx/log2=log(x+6)/(log2^2) => 2*logx=log(x+6)=> x^2=x+6=> x^2-x-6=0=> (x+2)(x-3)=0=> x_1=3,x_2=-2#

Only `x=3` is acceptable because `x>0`

Answer 2

We have to use the properties of logarithms.

Explanation:

`log _2x=log_4 (x+6)`
`=> ln x/ln 2= ln (x+6)/ln 4`
`=>ln x/ln 2 = ln (x+6)/ln(2^2)`
`=>ln x/cancel(ln2) = ln (x+6)/(2* cancel(ln2))`
`=>2 ln x= ln(x+6)`
`=> ln (x^2) = ln(x+6)`
`=> x^2=x+6`
`=> x^2-x-6=0`

which is thus reduced to a quadratic equation which can be solved by the quadratic formula to get the two roots: `x=-2 and x=3`

For `x=3`, the arguments within the logarithms in the original equation come out to be positive, thus `x=3` is a valid solution.

But for `x=-2`, one of the logarithmic argument is `-2` , giving rise to `log_2(-2)` which is completely meaningless. Thus `x=-2` is discarded and cannot be considered as a solution.

Answer 3

Make both sides have a base of `4`.

Explanation:

Change to `4^(log_2(x))=4^(log_4(x+6)`.

Notice that on the right side, the `4` and `log_4` will cancel, leaving just `(x+6)`.
`4^(log_2(x))=x+6`

Change the `4` to `2^2`.
`(2^2)^(log_2(x))=x+6`
`2^(2log_2(x))=x+6`

Remember this logarithm rule.
`2^(log_2(x^2))=x+6`

Like before, the `2` and `log_2` will cancel.
`x^2=x+6`
`x^2-x-6=0`
`(x-3)(x+2)=0`
`x=3 or cancel(x=-2`

Remember that in `log_a(b),b>0`.