Answer: \(\sqrt y=2.5\log_2(x+1)+5.4\); \(y=\frac{5929}{25}\); \(x=2^{-4/25}-1\approx-0.105\).

The key point is to rewrite the equation using logarithm laws or a suitable substitution.

Let

\(X=\log_2(x+1),\qquad Y=\sqrt y.\)

The straight line passes through \((2,10.4)\) and \((4,15.4)\). Its gradient is

\(\frac{15.4-10.4}{4-2}=\frac5{2}=2.5.\)

So

\(Y=2.5X+c.\)

Use \((2,10.4)\):

\(10.4=2.5(2)+c.\)

Thus

\(c=5.4.\)

Therefore

\(\sqrt y=2.5\log_2(x+1)+5.4.\)

When \(x=15\),

\(\log_2(x+1)=\log_2 16=4.\)

So

\(\sqrt y=2.5(4)+5.4=15.4.\)

Hence

When \(y=25\), \(\sqrt y=5\). Therefore

\(5=2.5\log_2(x+1)+5.4.\)

So

\(2.5\log_2(x+1)=-0.4,\)

and hence

\(\log_2(x+1)=-0.16=-\frac4{25}.\)

Therefore

\(x+1=2^{-4/25},\)

so

\(x=2^{-4/25}-1\approx -0.105.\)