# log 2 (2x+8)= 5 the 2 is under the log

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log 2 (2x + 8) =5

We know that if log x (y) = z , Then x^z= y

==> (2x+8) = 2^5

==> 2x + 8= 32

Now substract 8

==> 2x= 32-8= 24

Now divide by 2:

==> x= 12

To check:

log 2 (2(12)+8) = 5

log 2 ( 32) =5

log 2 (2^5) = 5

5 log2 (2) =5

5 (1) = 5

5=5

We'll write 5 as a result of using power property of logarithms:

5 = 5*1 = 5*log 2 (2) = log 2 (2^5)

Now, we'll re-write the equation:

log 2 (2x+8) = log 2 (2^5)

We'll use one to one property:

2x+8=32

We'll isolate 2x to the left side:

2x = 32-8

2x = 24

We'll divide by 2:

**x = 12**

**The solution of the equation is x=12.**

To solve log2(2x+8) = 5

Sollution:

log2(2x+8) = 5. Taking antilog,

2(2x+8) = 10^5.

4x+16 =10^5

4x = 10^5-16 = 99984

4x/4 = 99984/4 = 24996.

x = 24996.

If 2 is the base of log then

log (base2)(2x+8) = 5 = log( base2) 2^5. Taking antilog,

2x+8 = 2^5 =32.

2x=32-8 =24

2x/2=24/2 =12

x=12