Solve for x log2 x + log4 x + log8 x =11/6
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have to solve for x: log2 x + log4 x + log8 x =11/6
log (a) b = 1/ log (b) a
log2 x + log4 x + log8 x =11/6
=> 1 /...
(The entire section contains 63 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- x^log2 x + 8*x^-log2 x = 6. What is x?
- 1 Educator Answer
- logarithmsWhat is x if (log2 x)^2+log2 (4x)=4?
- 1 Educator Answer
- Solve for x x^(lnx/ln2)=6-8/x^(lnx/ln2)
- 1 Educator Answer
- Solve for x and y x^3-y^3=7 x^2+xy+y^2=7
- 2 Educator Answers
First, we'll impose the constraints of existence of logarithms:
x>0
Now, we'll change the bases of logarithms to base 2:
log2 x = log4 x*log2 4
log4 x = log2 x/log2 (2^2)
log4 x = log2 x/2log2 2
But log2 2 = 1
log4 x = log2 x/2
log2 x = log8 x*log2 8
log8 x = log2 x/3log2 2
log8 x = log2 x/3
We'll re-write the equation
log2 x + (log2 x)/2 + (log2 x)/3 = 11/6
We'll multiply by 6 both sides:
6 log2 x + 3 log2 x + 2 log2 x = 11
We'll add like terms:
11 log2 x = 11
We'll divide by 11:
log2 x = 1
x = 2^1
x = 2
Since 2 belongs to the interval of admissible values, we'll validate it as solution of the equation:
x = 2
Student Answers