6^log of base 6 is 5 +3^log of base 9 is 16 ????? pls give me solution for this?

Question

Answer

Solution to 6^log of base 6 is 5 and 3^log of base 9 is 16

Finding the Value of Logarithm

To solve this problem, we need to first understand the concept of logarithm. A logarithm is the power to which a number must be raised to get another number. In other words, it tells us what exponent we need to use to get a certain number.

For example, if we have the equation:

2^x = 8

We can rewrite this as:

x = log2 (8)

Here, log2 (8) means the logarithm of 8 with base 2. In other words, it tells us what power of 2 we need to use to get 8. In this case, it is 3, since 2^3 = 8.

Now, let's apply this concept to the given equations:

6^log6 (x) = 5

3^log9 (y) = 16

Here, log6 (x) means the logarithm of x with base 6. Similarly, log9 (y) means the logarithm of y with base 9.

Our goal is to find the value of x and y that satisfy these equations. To do this, we need to first find the values of log6 (x) and log9 (y) using the given information.

Using the Laws of Logarithm

To find the values of log6 (x) and log9 (y), we can use the laws of logarithm. These laws tell us how to simplify logarithmic expressions.

In particular, we can use the following laws:

logb (x * y) = logb (x) + logb (y)

logb (x / y) = logb (x) - logb (y)

logb (x^y) = y * logb (x)

Using these laws, we can simplify the given equations as follows:

6^log6 (x) = 5

log6 (x) * log6 (6) = log6 (5)

log6 (x) = log6 (5) / log6 (6)

log9 (y) * log3 (3) = log9 (16)

log9 (y) * 1 = log9 (16) / log9 (3)

log9 (y) = log9 (16) / log9 (3)

Calculating the Value of Logarithm

Now that we have simplified the equations, we can calculate the values of log6 (x) and log9 (y) using a calculator or by hand using the change of base formula.

Using a

Answer

By the property a^log of base a is b=bso 6^log of base 6 is 5=5....oknow 3^logbase9 is 16hm..yha kuch krna pdega ...ha we ll do tho according to some proprty jha tk mujhe pta hi ☆☆log base a is c=log base b is c/log base b is a☆☆so..yha pe 3^me hi na tho we ll take b =3..ok to eliminate it..log base 9 is 16=log base 3 is 16/log base 3 is 9=logbase3 is16/2logbase 3is 3((logm^n=n logm))=log base 3is 16/2 )((logbase a is a=1)=log base 3 is 16^1/2((loga)/b=log a^1/b))=log base 3is 4 now ....log base 9 is 16 =log base3 is 43^log base 9 is 16 =3^log 3is 4=4((by the proprty a^log base a is b=b))Finally ans is 5+4=9...Hope u got it...if not tell me ill expain again..

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