2log3x=x−−√+1You may only verify that x=9 is a roots as
x=9 is a roots as2log39=22=4
x=9 is a roots as2log39=22=4and
4=9–√+1.
Below is an algebraic derivation. Reexpress both sides of the equation,
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1as
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3x
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3x
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3x
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3xSince 4=3+1, the exponent must be one, i.e.
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3xSince 4=3+1, the exponent must be one, i.e.12log3x=1
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3xSince 4=3+1, the exponent must be one, i.e.12log3x=1which leads to the solution,
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3xSince 4=3+1, the exponent must be one, i.e.12log3x=1which leads to the solution,x=9
Answers & Comments
Verified answer
Step-by-step explanation:
Note that
alogbc=clogba
Sothe equation becomes
2log3x=x−−√+1
You may only verify that x=9 is a roots as
2log39=22=4
and
4=9–√+1.
Answer :
Note that
alogbc = clogba
clogbaSothe equation becomes
2log3x=x−−√+1
2log3x=x−−√+1You may only verify that x=9 is a roots as
x=9 is a roots as2log39=22=4
x=9 is a roots as2log39=22=4and
4=9–√+1.
Below is an algebraic derivation. Reexpress both sides of the equation,
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1as
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3x
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3x
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3x
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3xSince 4=3+1, the exponent must be one, i.e.
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3xSince 4=3+1, the exponent must be one, i.e.12log3x=1
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3xSince 4=3+1, the exponent must be one, i.e.12log3x=1which leads to the solution,
Below is an algebraic derivation. Reexpress both sides of the equation,xlog32=x−−√+1asRHS=x−−√+1=3log3x√+1=312log3x+112log3xLHS=xlog32=(3log3x)log32=(3log32)log3x=2log3x=412log3xAs a result, the original equation can be expressed in a form of identical exponents,412log3x=312log3x+112log3xSince 4=3+1, the exponent must be one, i.e.12log3x=1which leads to the solution,x=9