What is log(base 2) 32?

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Let `log_2(32)=x`

Since `log_ax=n` means that `a^n=x` we can write:

`2^x=32`

`rArr 2^x=2^5`

Since, the bases in the above equation are equal i.e 2, we can set the exponents equal. So,

`x=5`

Therefore, the value of `log_2(32)` is **5.**

**Sources:**

To determine the value of `log_(2)32`

we use the meaning of logarithms:`log_(a)b = x rArr a^x = b`

Therefore we let `log_(2)32 = x`` `

We rewrite as `2^x = 32`

Next, we'll rewrite 32 as a power of 2. 32 = `2^5` ` `

`2^x = 2^5`

Therefore **x = 5** as we have the same base

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