已知 $u = 2^x$，将 $2^{x+1} = 5(2^x) + 3$ 表示为关于 $u$ 的方程。 由此求出使得 $2^{x+1} = 5(2^x) + 3$ 成立的 $x$ 的值，结果保留 2 位有效数字。

1. Rewrite the term $2^{x+1}$ using laws of indices

$2^{x+1} = 2^x \cdot 2^1 = 2(2^x)$

2. Substitute $u = 2^x$ into the equation

$2u = 5u + 3$

1. Write down the quadratic equation in $u$ from part (a) and factorise it

From part (a): $2u^2 - 5u - 3 = 0$.

Factorising: $(2u + 1)(u - 3) = 0$

so $u = -\dfrac{1}{2}$ or $u = 3$.

2. Reject the invalid root

Since $u = 2^x > 0$ for all real $x$, the value $u = -\dfrac{1}{2}$ is rejected.

Hence $u = 3$, i.e. $2^x = 3$.

3. Solve for $x$ using logarithms

$2^x = 3$

$x = \log_2 3 = \dfrac{\ln 3}{\ln 2}$

4. Evaluate and round to 2 significant figures

$x = 1.5849\ldots$

$x \approx 1.6$ (2 s.f.)