Can anybody please solve it.question is related to logarithms, range?

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1 Answer
Apr 27, 2018

0<=x<=1 and x=4 are all solutions to this equation.

Explanation:

First note that by using the change of base formula

log_9x=(log_3x)/(log_3"9")=(log_3x)/2.

Therefore

2log_9x=log_3x, and

log_3x=log_9x^2.

So we can transform

log_3(sqrt(x)+abs(sqrt(x)-1))=log_9(4sqrt(x)-3+4abs(sqrt(x)-1))

to

log_9(sqrt(x)+abs(sqrt(x)-1))^2=log_9(4sqrt(x)-3+4abs(sqrt(x)-1)).

Since the logarithm is a function, if the logs are equal then their arguments are equal.

(sqrt(x)+abs(sqrt(x)-1))^2=4sqrt(x)-3+4abs(sqrt(x)-1)

Because of the absolute value expressions, there are two cases here - when x>=1 and when 0<=x<1.

When x>=1 we can ignore the absolute value signs and write

(2sqrt(x)-1)^2=4sqrt(x)-3+4(sqrt(x)-1)

4x-4sqrt(x)+1=8sqrt(x)-7

4x-12sqrt(x)+8=0

x-3sqrt(x)+2=0

(sqrt(x)-2)(sqrt(x)-1)=0

sqrt(x)=2 and sqrt(x)=1 which means that x=4, and x=1 are both solutions. In fact, if we plug these values for x back into the original equation, we can see that both of these values work.

For x=1

log_3(sqrt(1)+abs(sqrt(1)-1))=log_9(4sqrt(1)-3+4abs(sqrt(1)-1))

log_3(1)=log_9(1)

0=0.

For x=4

log_3(sqrt(4)+abs(sqrt(4)-1))=log_9(4sqrt(4)-3+4abs(sqrt(4)-1))

log_3(3)=log_9(9)

1=1.

Now for when 0<=x<1, we can rewrite the equation as

(sqrt(x)-sqrt(x)+1))^2=4sqrt(x)-3+4(-sqrt(x)+1)

1=1

Because this is ALWAYS true, there are an infinite number of solutions between and including 0 and 1.