This is an interesting question which I came across under **Additional Mathematics (A-Math)**:

Find the value of *x*.

Who is courageous to work on this question?

I will respond to this question when someone discusses about this question first :)

**Update**:

**(Apply Power Law)**

**(multiply by 4 on both sides of equal sign)**

**(Apply Power Law)**

**(Apply Quotient Law)**

**(Factorise)**

AGK says

(log-baseX-root3)(log-baseX-root8)=1.5(log-baseX-1/root2)

(0.5log-baseX-3)(1.5log-baseX-2)=1.5(log-baseX-2^-0.5)

[cancel 1.5 from both sides]

(0.5log-baseX-3)(log-baseX-2)= -0.5(log-baseX-2)

[cancel 0.5 from both sides]

(log-baseX-3)(log-baseX-2) + (log-baseX-2)= 0

(log-baseX-2) (log-baseX-3 + 1) = 0

therefore,

(log-baseX-2)=0 or (log-baseX-3 + 1)=0

X^0= 2 (rejected) (log-baseX-3)= -1

X^(-1) = 3

thus X= 1/3

hope i didnt make any mistakes

Logarithm rocks! XDD

alwaysLovely says

awesome!

Tks for the effort.

ahm97sic says

This question is fun. Perhaps, you will like to let your students to try to solve the question.

Solve the equation

(2/x)^(log n 4) - (3/x)^(log n 9) = 0

Regards,

ahm97sic

ahm97sic says

This is another interesting question. Perhaps, you will like to let your students to try to solve the question.

Solve (log a X)^(log b X) = X

where a, b are positive real numbers except 1, leave

the answer in terms of a and b.

Regards,

ahm97sic

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