## anonymous one year ago Which of the following logarithmic equations has no solution? log4(x + 2) = log4(6 - x) log4(-x) = 1 log4(x + 2) = 1 log4(x+2) = log4(x+6)

1. Nnesha

try to solve for x hint:hint: $\huge\rm log_b a=\log_b c$ $\large\rm \cancel{log_b} a=\cancel{\log_b }c$ a=c

2. Nnesha

bases of the log are same both sides so you cancel out log_B

3. anonymous

oh so it would just be x+2=6-x

4. welshfella

also try finding the log of -4 on your calculator

5. welshfella

- or the log of any negative for that matter

6. Nnesha

for 2nd) 3rd) you need to convert log to exponential form |dw:1439137679880:dw| after that you can plugin the x value to check ur answer

7. anonymous

non real answer, so that means it would be log4(-x) = 1 , right?

8. anonymous

so like a first step i think you need to know where is deffined the logaritmic function ? ok ?

9. welshfella

oh second thoughts x could be negative - in which case -x is positive - so its not the second option

10. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @arnavmishra oh so it would just be x+2=6-x $$\color{blue}{\text{End of Quote}}$$ yes right but you can solve for x there is a solution.

11. anonymous

i ended up with log4(x + 2) = 1, is that the right answer?

12. Nnesha

well solve that for x x= ??

13. Nnesha

for example first one $x+2=6-x$ solve for x move the -x to the left side and 2 to the right side $x+x=6-2$$2x=4$ x=2 now plugin the 2 into the original equation for x $\log_4(2+2)=\log_4(6-2)$ both sides are equal so 2 is a solution

14. phi

the last choice $\log_4(x+2) = \log_4(x+6)$ make each side the exponent of the base 4: $4^{\log_4(x+2) }= 4^{\log_4(x+6)}\\ x+2= x+6$ (we could have "jumped" to this equation) now add -x to both sides: $x-x+2 = x-x + 6\\ 2= 6$ we know 2 is not 6, so something is wrong... namely there is no solution to the original expression.