Question 141333
Please help me solve this simultaneous equation:

{{{log(3,x)=y=log(9,(2x-1))}}}
<pre><font size = 4 color = "indigo"><b>
Write two separate equations:

{{{y=log(3,x)}}}
{{{y=log(9,(2x-1))}}}

Use this principle to rewrite each: 
                {{{A=log(B,C)}}} can be rewritten as {{{C=B^A}}}

Rewriting the first:
{{{x=3^y}}}

Rewriting the second:
{{{2x-1=9^y}}}

Now rewrite {{{9}}} as {{{(3*3)}}}

{{{2x-1=(3*3)^y}}}

or, multiplying exponents:

{{{2x-1=(3^y)(3^y)}}}

So now we have this system:

{{{2x-1=(3^y)(3^y)}}}
{{{x=3^y}}}

Using the second, we can substitute
{{{x}}} for {{{3^y}}} in the first:

{{{2x-1=(x)(x)}}}

{{{2x-1=x^2}}}

Get 0 on the left

{{{0=x^2-2x+1}}}

Factor the right side:

{{{0=(x-1)(x-1)}}}

Using the zero factor principle,

{{{x-1=0}}} gives {{{x=1}}}
{{{x-1=0}}} gives {{{x=1}}}

So there is only one value for {{{x}}},
which is {{{1}}}

Substitute {{{1}}} for x in {{{x=3^y}}}

{{{1=3^y}}}

Now we use the rule:
            Any number raised to the zero power, 
            except zero itself, equals 1

Thus {{{y=0}}}

and the solution is {{{x=1}}}, {{{y=0}}}

Let's check:

{{{log(3,x)=y=log(9,(2x-1))}}}
{{{log(3,(1))=0=log(9,(2(1)-1))}}}
{{{log(3,(1))=0=log(9,(2-1))}}}
{{{log(3,(1))=0=log(9,(1))}}}
{{{log(3,(1))=0}}} can be rewritten as {{{3^0=1}}} which is true.
{{{log(9,(1))=0}}} can be rewritten as {{{9^0=1}}} which is also true.

-------------------------------

and another one:

{{{log((x+y)) = 0}}}
{{{2*log(x) = log ((y+1))}}}

When no base is written, the base {{{10}}} is understood:

{{{log(10,(x+y)) = 0}}}
{{{2*log(10,(x)) = log( 10,(y+1) )}}}

We can rewrite the first using the rule:
         {{{log(B,A)=C}}} can be rewritten as {{{A=B^C}}}
         
{{{log(10,(x+y)) = 0}}}
becomes
{{{x+y = 10^0}}}
{{{x+y=1}}}

We can use this rule on the left side of the second eq:
                      {{{A*log(B,(C))=log(B,(C^A))}}}

{{{2*log(10,(x)) = log( 10,(y+1) )}}}
becomes
{{{log(10,(x^2)) = log( 10,(y+1) )}}}

Now we use the principle: 
                {{{log(B,A)=log(B,C)}}} can be rewritten {{{A=C}}}

{{{x^2=y+1}}}

So we have the system of ewquations:

{{{x+y=1}}}
{{{x^2=y+1}}}
          
Can you solve that system of equations by substitution?
If not post again asking how.

That last system of equations has two ordered pairs 
of solutions:

({{{x}}},{{{y}}})= ({{{1}}},{{{0}}}) and ({{{x}}},{{{y}}}) = ({{{-2}}},{{{3}}})

However we must check them, because sometimes a solution
to our final equations is not a solution to the original
equation:

Checking ({{{x}}},{{{y}}})= ({{{1}}},{{{0}}}) in the first equation:

{{{log(10,(1+0)) = 0}}}
{{{log(10,(1)) = 0}}}
{{{10^0=1}}}
{{{1=1}}}

Checking ({{{x}}},{{{y}}})= ({{{1}}},{{{0}}}) in the second equation:

{{{2*log(10,(1)) = log( 10,(0+1) )}}}

{{{2*log(10,1) = log( 10,(0+1) )}}}
{{{2*log(10,1) = log( 10,(1) )}}}
Since {{{log(10,1)=0}}}
{{{2*0=0}}}
{{{0=0}}}

----

Checking ({{{x}}},{{{y}}})= ({{{-2}}},{{{3}}}) in the first equation:

{{{log(10,(-2+3)) = 0}}}
{{{log(10,(1)) = 0}}}
{{{10^0=1}}}
{{{1=1}}}

Checking ({{{x}}},{{{y}}})= ({{{-2}}},{{{3}}}) in the second equation:

{{{2*log(10,(-2)) = log( 10,(3-2) )}}}

We can stop here because the first term is 
undefined because the logarithm of a negative 
number is not defined (except in certain 
advanced mathematics, but never in
ordinary algebra)

So there is but one solution, ({{{x}}},{{{y}}})= ({{{1}}},{{{0}}})

Edwin</pre>