Question

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${x}^{ log_{3}2} = \sqrt{x} + 1$
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Answer 1

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 2

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