Target Mathematics

1.51: Logarithmic equations

4PM1/2/June/2015 — Question 10 · 11 marks

(a) Find the value of log⁡39\log_3 9. (1)
Given that log⁡94=klog⁡34\log_9 4=k\log_3 4,
(b) find the value of kk. (2)
(c) Show that
2xlog⁡3x−3log⁡3x−4xlog⁡94+6log⁡94=log⁡3(x4)2x−32x\log_3x-3\log_3x-4x\log_9 4+6\log_9 4=\log_3\left(\dfrac{x}{4}\right)^{2x-3}
(5)
(d) Hence solve the equation 2xlog⁡3x−3log⁡3x−4xlog⁡94+6log⁡94=02x\log_3x-3\log_3x-4x\log_9 4+6\log_9 4=0. (3)