If `4^44 + 4^44 + 4^44 + 4^44 = 4^x` , then x equals:a.45b.203c.4444d.44444

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Wilson2014's profile pic

Posted on

One important logarithmic property that can be used to solve this:

`log x^a = a*log x`

 

`4^44+4^44+4^44+4^44=4^x`

Taking the log of both sides:

`log (4^44+4^44+4^44+4^44) = log (4^x)`

`=> log (4^44+4^44+4^44+4^44) = x* log (4)`

`=> (log(4^44+4^44+4^44+4^44))/(log(4)) = x`

As explained in the other answer, the sum of 4 to the 44th power equals 4 to the 45th power.

`=> (log (4^45))/(log(4)) = x`

`(45*log(4))/(log(4)) = x`  

The logs cancel out. Therefore `45=x` .  

mjripalda's profile pic

Posted on

`4^44+4^44+4^44+4^44=4^x`

Since 4^44 is added four times by itself, left side can be express as:

`4*4^44=4^x`

To simplify the left side further, apply the properties of exponent when multiplying same base which is `a^m*a^n=a^(m+n)` .

`4^1*4^44=4^x`

`4^(44+1)=4^x`

`4^45=4^x`

Since both sides of the equation has same base, to solve for x, set the exponents of each side equal to each other.

`45=x`

Hence, x=45.

madmoni06's profile pic

Posted on

Can anyone plz solve it by using logarithm.....

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