Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. (If there is no solution, enter NO SOLUTION.)
log2X+log2(x-3)=2

Answer 1

\[\log_{2}x+\log_{2}(x-3)=2 \]

Answer 2

sorry about weird looking format on initial question.. The above is the correct format

Answer 3

so do you know the property of logarithms ?

Answer 4

how you can simplifie it

log A +log B = ?

Answer 5

using property of logarithms

Answer 6

so than you need to learn it again than

log a + log b = log a*b

ok ?

Answer 7

suppose being base of 2

Answer 8

so and using log base 2 how you can writing the number 2 ?

Answer 9

would it be \[2\ln +2\ln (x-3)=2\] to start ?

Answer 10

not is necessary because there is speciffied that us logarithm base 2

Answer 11

\[2\log_{2}x+2\log_{2}(x-3)=2 \] is that correct to start process?

Answer 12

not is

so there is


log2 x + log2 (x-3) =2

yes ?

Answer 13

using logarithm base 2 you can writing the 2 how ?

Answer 14

so like 2= log2 2^2

right ?

Answer 15

than because you know that log2 a + log2 b = log2 a*b so using this property
your exercise will be

log2 x +log2 (x-3) =log2 2^2
log2 (x*(x-3)) = log2 2^2 so than result
x*(x-3)=2^2
x^2 -3x =4
x^2 -3x -4=0 factorized will get
(x-4)(x+1)=0 so than

x-4=0
x=4
and
x-1=0
x=1
so but you know that the function of logarithm is defined just in interval of (0,+infinite) so from this result that the value x=1 not is right because there is log2 (x-3) where using x=1 you will get log2 of -2 what is undefined
so than will be right just x=4

hope so much that i have wrote it right,correct and understandably easy right

good luck

bye