Solve for n in (log n)2 = log2 n log2 8 Make x the subject of the formula 2x+r 4 In the expansion of (a-b/9 2), the term independent of a is 735. Find the value of b.

Mlinganyo uliotolewa ni:

$$(\log n)^2 = \log_2 n - \log_2 8$$

Katika muktadha wa maswali ya hisabati ya KCSE, $\log n$ bila base maalum inamaanisha $\log_{10} n$. Hivyo, mlinganyo unakuwa:

$$(\log_{10} n)^2 = \log_2 n - \log_2 8$$

Tunajua kuwa $8 = 2^3$. Hivyo,

$$\log_2 8 = \log_2 (2^3) = 3$$

Mlinganyo unakuwa:

$$(\log_{10} n)^2 = \log_2 n - 3$$

Tumia formula ya kubadilisha base: $\log_b a = \frac{\log_c a}{\log_c b}$.
Hivyo, $\log_2 n = \frac{\log_{10} n}{\log_{10} 2}$.
Mlinganyo unakuwa:

$$(\log_{10} n)^2 = \frac{\log_{10} n}{\log_{10} 2} - 3$$

Badilisha $x$ kwenye mlinganyo:

$$x^2 = \frac{x}{\log_{10} 2} - 3$$

Panga upya mlinganyo ili uwe katika mfumo wa kawaida wa quadratic $ax^2 + bx + c = 0$:

$$x^2 - \frac{1}{\log_{10} 2} x + 3 = 0$$

Kwanza, hesabu thamani ya $\log_{10} 2$:
$\log_{10} 2 \approx 0.30103$
Sasa, badilisha thamani hii kwenye mlinganyo:

$$x^2 - \frac{1}{0.30103} x + 3 = 0$$ $$x^2 - 3.3219 x + 3 = 0$$

Tumia formula ya quadratic: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Katika mlinganyo huu, $a=1$, $b=-3.3219$, na $c=3$.
Hesabu kibaguzi (discriminant), $\Delta = b^2 - 4ac$:

$$\Delta = (-3.3219)^2 - 4(1)(3)$$ $$\Delta = 11.0350 - 12$$ $$\Delta = -0.9650$$

Kwa kuwa kibaguzi ($\Delta$) ni hasi (chini ya sifuri), hakuna suluhisho halisi (real solutions) kwa $x$. Hii inamaanisha kuwa hakuna thamani halisi ya $\log_{10} n$ inayoweza kukidhi mlinganyo huu, na kwa hivyo, hakuna thamani halisi ya $n$.

Hakuna suluhisho halisi kwa $n$.

$$\boxed{\text{Hakuna suluhisho halisi kwa } n}$$