Solve unknown variable in the following: a) logm 9 + logm (m2) = 5/2, (d) log3 z + logz 3 = 4; (b) log2 y = log4 (y+6), (e) log3 4 = ln x + ln 4; (c) ln(ln(3x)) = 0, (f) 4x − 2 = x, e^x

Tumia kanuni ya kubadilisha msingi wa logariti: $\log_b a = \frac{\log_c a}{\log_c b}$. Tutatumia $c=m$.

$$\log_{(m^2)} 3 = \frac{\log_m 3}{\log_m (m^2)}$$

Tunajua kuwa $\log_m (m^2) = 2 \log_m m = 2 \times 1 = 2$.
Hivyo,

$$\log_{(m^2)} 3 = \frac{\log_m 3}{2}$$

Mlinganyo wetu unakuwa:

$$\log_m 9 + \frac{\log_m 3}{2} = \frac{5}{2}$$

Tunajua $9 = 3^2$. Tumia kanuni $\log_b (a^n) = n \log_b a$.

$$\log_m 9 = \log_m (3^2) = 2 \log_m 3$$

Badilisha hii kwenye mlinganyo:

$$2 \log_m 3 + \frac{\log_m 3}{2} = \frac{5}{2}$$

Wacha $X = \log_m 3$. Mlinganyo unakuwa:

$$2X + \frac{X}{2} = \frac{5}{2}$$

Ili kuondoa sehemu, zidisha pande zote mbili kwa 2:

$$2(2X) + 2\left(\frac{X}{2}\right) = 2\left(\frac{5}{2}\right)$$ $$4X + X = 5$$ $$5X = 5$$ $$X = 1$$

Hivyo, $\log_m 3 = 1$.

Kutoka kwa ufafanuzi wa logariti, ikiwa $\log_b a = c$, basi $b^c = a$.
Hivyo, $m^1 = 3$.

$$m = 3$$

Kwa kuwa $m=3$ inakidhi masharti ya msingi wa logariti ($m>0$ na $m \neq 1$), suluhisho ni halali.

Thamani ya $m$ ni:

$$\boxed{m = 3}$$