Question

log(base4)(x+3)+log(base4)(2-x)=1

Answer 1

Solve for x?

Use rules of logs to combine on left side.

\[\log_{4}(x+3) +\log_{4}(2-x)=\log_{4}[(x+3) (2-x)]=\log_{4}[(x^2-x+6)]\]
So now you have
\[log_{4}[(x^2-x+6)]=1\]

Convert to exponential form:\[4^{1}=x^2-x+6\]\[x^2-x+6=4\]\[x^2-x+2=0\]

Now solve this quadratic equation. Beware of extraneous solutions.

Answer 2

\[\log_{4} \left( x+3 \right)+\log_{4} \left( 2-x \right)=1\]
\[\log_{4} \left( x+3 \right)\left( 2-x \right)=1\]
\[4^{1}=\left( x+3 \right)\left( 2-x \right)\]
now you can solve as above

Answer 3

thank you both! :)

Answer 4

\[4=2x-x ^{2}+6-3x orx ^{2}+x-2=0\]
\[x ^{2}+2x-x-2=0,\]
x(x+2)-1(x+2)=0
(x+2)(x-1)=0
either x=-2 or x=1

Answer 5

yw