# log(base2)x+log(base2)(x+5)=k?

solve for k in terms of k.

solve for x when if k=7

### 2 Answers

- SamwiseLv 71 decade agoFavorite Answer
"Solve for k in terms of k" doesn't make sense: k=k is obvious.

I assume one of those k's ought to have been an x.

Let's simplify the whole thing and see if we can get a cleaner statement of the relationship between x and k:

log(base2)x+log(base2)(x+5)=k

log(base2) [x(x+5)] = k {by properties of logarithms}

That's an expression for k in terms of x

x(x+5) = 2^k {definition of logarithms}

x^2 + 5x = 2^k

x^2 + 5x + 25/4 = 2^k + 25/4 {completing the square}

(x + 5/2)^2 = 2^k + 25/4

x + 5/2 = √(2^k + 25/4)

(because a negative value won't satisfy the logarithm)

x = - 5/2 + √(2^k + 25/4)

That's x in terms of k.

When k=7, we have

x - 5/2 + √(2^k + 25/4) = - 5/2 + √(128 + 25/4)

= - 5/2 + √(537/4) = (√537 - 5) / 2 = about 9.09

Using the equation

x^2 + 5x = 2^k

above, we see that with k=7,

x^2 + 5x - 128 = 0

and the solution above does, indeed, fit.