Solve ( l o g <mrow class="MJX-TeXAtom-ORD"> 2 </mrow> </msub

Averi Mitchell

Averi Mitchell

Answered question

2022-06-19

Solve ( l o g 2 ( x + 1 ) ) 2 = 4
( l o g 2 ( x + 1 ) ) 2 = 4
l o g 2 ( x + 1 ) l o g 2 ( x + 1 ) = l o g 2 16
x 2 + 2 x 15 = 0
( x + 1 ) ( x + 1 ) = 16
x 2 + 2 x + 1 = 16
x 2 + 2 x 15 = 0
( x + 5 ) ( x 3 ) = 0
x 1 = 5 ; x 2 = 3
The solution is only x 1 = 3. Is this correct?

Answer & Explanation

Kaydence Washington

Kaydence Washington

Beginner2022-06-20Added 32 answers

If
( log 2 ( x + 1 ) ) 2 = 4
then
log 2 ( x + 1 ) = + 2
or
log 2 ( x + 1 ) = 2
so x + 1 = 2 + 2 or x + 1 = 2 2
Lydia Carey

Lydia Carey

Beginner2022-06-21Added 9 answers

Your solution is incorrect. First note that
log 2 ( a 2 ) ( log 2 ( a ) ) 2
In your case, we have
( log 2 ( x + 1 ) ) 2 = 4 log 2 ( x + 1 ) = ± 2 x + 1 = 2 ± 2 x + 1 = 4  or  x + 1 = 1 4
Hence,
x = 3  or  x = 3 4

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