Solve the logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic expressions. Give the exact answer.

log4x2 = log4(5x + 14)

Question

Solve the logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic expressions. Give the exact answer.

log4x2 = log4(5x + 14)

mathstudent55 · Answer

If log x = log y, then x = y.

campbell_st · Answer

equate the expressions since the logs are equal and the same base

$$x^2 = 5x + 14$$


so solve the above expression


and remember you can't find the log of a number less than zero

mathstudent55 · Answer

^less than or equal to zero

anonymous · Answer

$$\log_4x^2 = \log_4(5x + 14) \implies x^2-5x-14=(x-7)(x+2)=0$$$$ \implies x=7~~ or~~ x=-2$$However, the second solution won't work, since -2 is not in the domain of the original problem, so the solution is $$x=7$$

mathstudent55 · Answer

@AnimalAin Why will x = -2 not work?

anonymous · Answer

Excellent point.  I guess I only get half credit on this one.....

anonymous · Answer

Yeah, it works fine.  Brain fade, I guess; after my bedtime.

mathstudent55 · Answer

$x = -2$

$ \log_{4} x^2 = \log_{4} (-2)^2  = \log_{4} 4 = 1 $

$ \log_{4} (5x + 14) = \log_{4} [5(-2) + 14]  = \log_{4} (-10 + 14) = 
\log_{4} 4 = 1$