Question

If a^2+b^2=2ab then loga

Answer 1

log24√2basex=2-log3√2 base x

Answer 2

To find log a, we can start by simplifying the equation a^2 + b^2 = 2ab.

Dividing both sides by ab gives (a^2 + b^2) / (ab) = 2.

Using the formula log (xy) = log x + log y, we can rewrite the equation as log (a^2) + log (b^2) - log (ab) = log 2.

Using the properties of logarithms, we can simplify further. log (a^2) + log (b^2) - log (ab) = log (a^2 * b^2 / ab) = log (ab).

Thus, log (ab) = log 2.

Taking the antilogarithm of both sides gives ab = 2.

Therefore, log a = log (ab) - log b = log 2 - log b.

Answer 3

To solve the equation log_2 (4√2) base x = 2 - log_3 (√2) base x, we can simplify each logarithm separately and then equate them.

First, let's simplify log_2 (4√2) base x.
Using the property of logarithms, log_a (b*c) = log_a (b) + log_a (c), we can split the logarithm as follows:
log_2 (4√2) base x = log_2 (4) base x + log_2 (√2) base x.

Since x is the base, log_2 (4) base x can be rewritten as:
log_2 (2^2) base x = 2 * log_2 (2) base x = 2.

Similarly, log_2 (√2) base x can be rewritten as:
log_2 (2^(1/2)) base x = (1/2) * log_2 (2) base x = 1/2.

Now let's simplify the other side of the equation, 2 - log_3 (√2) base x.
Using the property of logarithms, we can rewrite log_3 (√2) base x as:
log_3 (2^(1/2)) base x = (1/2) * log_3 (2) base x.

Now we can substitute the simplified values back into the equation:
2 - log_3 (√2) base x = 2 - (1/2) * log_3 (2) base x.

Now we have the equation:
2 = 2 - (1/2) * log_3 (2) base x.

To solve for log_3 (2) base x, we can subtract 2 from both sides:
0 = - (1/2) * log_3 (2) base x.

Now we have:
-(1/2) * log_3 (2) base x = 0.

Since the result of log_3 (2) base x is 0, we can conclude that:
2^0 base x = 3.

Therefore, x = 3.