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Loga^b logb^c = 0 then?
Most Upvoted Answer
Loga^b logb^c = 0 then?
Explanation:


The given equation is:


logab × logbc = 0


Rule of logarithms:


If logab = x, then ax = b


Using the above rule of logarithms:


logab × logbc = 0


If logab = x, then ax = b


If logbc = y, then by = c


Therefore, we can rewrite the given equation as:


ax × by = 1


Since ax and by are both positive, their product can only be equal to 1 if both ax and by are equal to 1.


Using the above conclusion:


Therefore, we have:


ax = 1 and by = 1


If ax = 1, then x = 0 (since a0 = 1 for any positive value of a)


If by = 1, then y = 0 (since b0 = 1 for any positive value of b)


Therefore, we have:


logab = 0 and logbc = 0


If logab = 0, then b = a0 = 1


If logbc = 0, then c = b0 = 1


Therefore, we have:


b = 1 and c = 1


Substituting these values in the original equation, we get:


loga1 × log11 = 0


Since loga1 = 0 (for any positive value of a) and log11 = 0, we have:


0 × 0 = 0


Conclusion:


Therefore, the given equation logab × logbc = 0 is true if and only if b = 1 and c = 1.
Community Answer
Loga^b logb^c = 0 then?
If the equation is loga^b * logb^c = 0, we can simplify it using the properties of logarithms.

First, let's recall the following properties:

loga^b = c if and only if a^c = b.
loga^b * logb^c = loga^b^c.
Using property 1, we can rewrite the equation as:

(a^c) * (b^c) = 1.

Now, let's analyze the possibilities:

If a^c = 1 and b^c = 1, then the equation is satisfied.

If a^c ≠ 1 and b^c = 1, then the equation is not satisfied because (a^c) * (b^c) ≠ 1.

If a^c = 1 and b^c ≠ 1, then the equation is not satisfied because (a^c) * (b^c) ≠ 1.

If a^c ≠ 1 and b^c ≠ 1, then the equation is not satisfied because (a^c) * (b^c) ≠ 1.

Therefore, the only possibility for the equation loga^b * logb^c = 0 to be true is if a^c = 1 and b^c = 1.
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Loga^b logb^c = 0 then?
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