# If 10^a=5, 10^b=6, what is the value of 10^(2a+b). Please use log.

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### 3 Answers

It is given that 10^a = 5 and 10^b = 6. I do not see any reason why logarithms have to be used to find 10^(2a + b).

But as you would like it done that way let's see how it can be done using logs.

10^a = 5

take the log to the base 10 of both the sides

=> log 10^a = log 5

use the property log a^b = b*log a

=> a * log 10 = log 5

the log of the base is equal to 1

=> a = log 5

Similarly, 10^b = 6 gives b = log 6

Now if we take the log of 10^(2a + b) we get

log 10^(2a + b) = 2a + b

=> 2*log 5 + log 6

=> log 25 + log 6

use the property log a + log b = log a*b

=> log 25*6

=> log 150

As log 10^(2a + b) = log 150

=> 10^(2a + b) = 150

**We arrive at the result that 10^(2a + b) = 150.**

By using log.

We will rewrite exponent into logarithm forms.

10^a = 5 ==> log 5 = a...........(1)

10^b = 6 ==> log 6 = b...........(2)

Now we will find the value of 10^(2a+b)

==> Let 10^(2a+b) = x

==> we will take log to both sides.

==> log 10^(2a+b) = log x

==> 2a+ b= log x

==> log x = 2*log 5 + log 6

==> Now we will use logarithm properties to solve.

==> We know that alog b = log b^a

==> log x = log 5^2 + log 6

==> log x = log 25 + log 6

Now we know that log a + log b = log (ab)

==> log x = log 25*6

==> log x = log 150

==> x = 150

==>** 10^(2a+b) = 150**

10^a = 5 ==> log 5 = a...........(1)

10^b= 6 ==> log 6 = b...............(2)

We need to find the value of 10^(2a+b)

We will use exponent properties to find the values.

We know that x^(a+b) = x^a * x^b

==> 10^(2a+b) = 10^2a * 10^b

Now we also know that x^(ab) = (x^a)^b

==> 10^(2a+b) = (10^a)^2 * 10^b

Now we will substitute with the values of (1) and (2).

==> 10^(2a+b) = 5^2 * 6 = 25 * 6 = 150

**==> 10^(2a+b) = 150.**