Question

How do i solve the rest of this logarithmic equation log(subscript16)32=x+2. So far I have 32=16^(x+2) x+2=log(subscript16)32. Don't know how to solve from here can someone please help me?

Answer 1

I think x=-0.75 now I am checking my work

Answer 2

\(\color{#000000 }{ \displaystyle16^{x+2}=32 }\)

\(\color{#000000 }{ \displaystyle \log_{16}16^{x+2}=\log_{16}32 }\)

\(\color{#000000 }{ \displaystyle (x+2)\log_{16}16=\log_{16}32 }\)

\(\color{#000000 }{ \displaystyle (x+2)\cdot 1=\log_{16}32 }\)

\(\color{#000000 }{ \displaystyle x+2=\log_{16}32 }\)

Answer 3

I am just confirming that your result for converting into logarithm is right

Answer 4

Really? Thanks!

Answer 5

you can further simplify it by applying the rules

\(\color{#000000 }{ \displaystyle x+2=\log_{16}32 }\)
\(\color{#000000 }{ \displaystyle x+2=\log_{16}(16\cdot 2) }\)
\(\color{#000000 }{ \displaystyle x+2=\log_{16}(16) +\log_{16}(2) }\)
\(\color{#000000 }{ \displaystyle x+2=1 +\log_{16}(2) }\)

Answer 6

\(\color{#000000 }{ \displaystyle x=-1+\log_{16}(2) }\)

Answer 7

and then use the fact that \(2^4=16\quad \Longrightarrow \sqrt[4]{16}\quad \Longrightarrow {16}^{1/4}\)

Answer 8

\(\color{#000000 }{ \displaystyle x=-1+\log_{16}(2) }\)

\(\color{#000000 }{ \displaystyle x=-1+\log_{16}(16^{1/4}) }\)

Answer 9

so your answer is right,
because after taking the exponent out you get exactly that

Answer 10

I understand that more

Answer 11

\(\color{#000000 }{ \displaystyle x=-1+\log_{16}(16^{1/4}) }\)

\(\color{#000000 }{ \displaystyle x=-1+(1/4)\log_{16}(16) }\)

\(\color{#000000 }{ \displaystyle x=-1+(1/4)\cdot 1}\)

\(\color{#000000 }{ \displaystyle x=-1+(1/4)}\)

\(\color{#000000 }{ \displaystyle x=-1+0.25=-0.75 }\)

Answer 12

good job\( : \)

Answer 13

Thank you!

Answer 14

Anytime!

Answer 15

\[16^{x+2}=32\]
\[\left( 2^4 \right)^{x+2}=2^5\]
\[2^{4\left( x+2 \right)}=2^5,4\left( x+2 \right)=5,4x=5-8=-3,x=-\frac{ 3 }{ 4 }\]