How to take the log of an equation?

Solving Log Equation 3*log(x-15)=(1/4)^x & log(28-x^3)/log(4-x)=7?

• Answer:

  1) 3Log(x - 15) = (¼)^x Graphing seems the only reasonable solution method although some deduction can (almost) get you there. (¼)^x will always be positive and can range from effectively 0 to +∞. Consequently, 3Log(x - 15) and thus Log(x - 15) must also be positive which means x >16. When x = 16, Log1 = 0 and thus 3Log1 = 0, (¼)^16 = 2.33 x 10^-10 When x = 16.1, 3Log1.1 = 0.124 whilst (¼)^16.1 = 2.03 x 10^-10 Thus, as x increases, 3Log(x - 15) increases whilst (¼)^x decreases in value. The solution is therefore that x ≅ 16 (marginally greater than 16) 2) Log(28 - x³)/Log(4 - x) = 7 Log(28 - x³) = 7Log(4 - x) Then (28 - x³) = (4 - x)^7.......which can be expanded and a polynomial equation of degree 7 solved to produce the answer. Graphing is much easier. One obvious solution occurs when x = 3 : Log(28 - 27) = Log1 = 0 : = : 7 Log(4 - 3) = 7Log1 = 0