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2^n+2^98=2^99 what does n =

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We are given the following equation:


2^n+2^(98)=2^(99)

We are asked to solve for "n". To do that we will first subtract 2 to the 98 from both sides:


2^n=2^(99)-2^(98)

Solving the operations we get:


2^n=3.2*10^(29)

Now, we take natural logarithm to both sides:


\ln 2^n=\ln 3.2*10^(29)

Now, we use the following property of logarithms:


\ln x^y=y\ln x

Applying the property we get:


n\ln 2=\ln 3.2*10^(29)

Now, we divide both sides by ln 2:


n=(\ln 3.2*10^(29))/(\ln 2)

Solving the operations we get:


n=98

Therefore, the value of "n" is 98.

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