How do you solve 2^x =128?

As we know 128 = 2^7, we don't have to use logarithm in this case,

2^x = 128

2^x = 2^7

x = 7

As we know 128 = 2^7, we don't have to use logarithm in this case,

2^x = 128

2^x = 2^7

x = 7

but in the harder cases, we take log/ln from both sides and use a calculator to find x:

2^x = 128

ln 2^x = ln 128

x * ln 2 = ln 128

x = ln 128 / ln 2

x = 7

Take the logarithm (any convenient base) of both sides. Since 128 = 2⁷, base 2 is natural

x lb2 = lb128

x = lb128/lb2 = 7

Note: lb(x) is one notation for the binary (base 2) logarithm.

Two ways, but we can do this the easy way..

1. Have the same base (notice 128 is equal to 2^7)

2^x = 2^7

x = 7

2. Done.

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EDIT: You can do this using logarithms, but I like this way because it's easy. The other way is...

2^x = 128

Take ln from both sides

ln 2^x = ln 128

x * ln 2 = ln 128 (x In 2 is the same as In 2^x)

Divide both sides by In 2

x = ln 128 / ln 2

x = 7

x=7; 2^7=128

yes this is a logarithm, it would be changed to look like: log(2)128=7

The (2) would be a subscript

This is the solution of a power function.

To solve, take logarithms of both sides, noting log(2^x) = x*log2

Divide by log 2

Should give the answer 7

2^x =128

2^x =2^7

x =7 answer//