What is the answer to log_{2}(x+1)=78 ?

The answer to log_{2}(x+1)=78 is x=2^{78}-1

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log_{2}(x+1)=78

lo g_{2} (x + 1) = 78

x = 2^{78} - 1

Decimal Notation

x = 3.02231 E 23

Solution steps

lo g_{2} (x + 1) = 78

Apply log rules

x + 1 = 2^{78}

Solve

x + 1 = 2^{78} : x = 2^{78} - 1

x = 2^{78} - 1

lo g_{4} (x^{2}) + 4 lo g_{4} (x) = 4

2 lo g_{10} (n) = lo g_{10} (n^{2})

lo g_{x} (5 \cdot x - 6) = 2

lo g_{3} (2 \cdot x) = 10

(2 x)^{l n (2 x)} = e^{2} (2 x)

What is the answer to log_{2}(x+1)=78 ?
The answer to log_{2}(x+1)=78 is x=2^{78}-1
Solve for x: log_{2}(x+1)=78
The solution is x=2^{78}-1