Question 644781
<pre>

This requires these two principles of exponents

(1)    {{{a^(b+c)}}} = {{{a^b*a^c}}}  <--The reversal of adding exponents

(2)    {{{(a^b)/(c^b)}}} = {{{(a/c)^b}}} <--The reversal of the power of a fraction 


{{{3^(2x+1)}}} - {{{5*2^(2x+1)}}} = {{{6^x}}}

Divide through by {{{2^(2x+1)}}}

{{{3^(2x+1)/2^(2x+1)}}} - 5 = {{{6^x/(2^(2x+1))}}}

Use principle (2) on the left side
Write 6 as 2*3 on the right

{{{(3/2)^(2x+1)}}} - 5 = {{{(2*3)^x/(2^(2x+1))}}}

Use principle (1) on the right side

{{{(3/2)^(2x+1)}}} - 5 = {{{(2^x*3^x)/(2^(2x+1))}}}

Use principle (1) on the left
Subtract exponents of 2 on the right

{{{(3/2)^(2x)*(3/2)^1)}}} - 5 = {{{3^x/(2^(2x+1-x))}}}

Simplify

{{{(3/2)^(2x)*(3/2))}}} - 5 = {{{3^x/(2^(x+1))}}}

Swap factors on the left, use principle (1) on the right denominator

{{{(3/2)(3/2)^(2x))}}} - 5 = {{{3^x/(2^x*2^1))}}}

Remove the 1 exponent

{{{(3/2)(3/2)^(2x))}}} - 5 = {{{3^x/(2^x*2))}}}

Multiply through by 2

{{{2*(3/2)(3/2)^(2x))}}} - 2·5 = {{{2*expr(3^x/(2^x*2)))}}}

{{{cross(2)*(3/cross(2))(3/2)^(2x))}}} - 10 = {{{cross(2)*expr(3^x/(2^x*cross(2))))}}}

{{{3(3/2)^(2x))}}} - 10 = {{{3^x/2^x}}}

{{{3(3/2)^(2x))}}} - 10 = {{{(3/2)^x}}}

{{{3(3/2)^(2x))}}} - {{{(3/2)^x}}} - 10 = 0

let u = {{{(3/2)^x}}}

Then uČ = {{{((3/2)^x)^2}}} = {{{(3/2)^(2x))}}}

Then we have

3uČ - u - 10 = 0

Factoring,

(u - 2)(3u + 5) = 0

Use the zero-factor principle:

u - 2 = 0;  3u + 5 = 0
    u = 2       3u = -5
                 u = {{{-5/3}}}

Use u = 2

Since u = {{{(3/2)^x}}}

{{{(3/2)^x}}} = 2

Take logs of both sides:

{{{log(((3/2)^x))}}} = {{{log((2))}}}

Use the principle {{{log((b^a))=a*log((b))}}}

{{{x*log((3/2))}}} = {{{log((2))}}}

Divide both sides by {{{log((3/2))}}}

{{{x}}} = {{{log((2))/log((3/2))}}} = 1.709511291

That's the only solution because if we try 

u = {{{-5/3}}}

{{{(3/2)^x}}} = {{{-5/3}}}

That cannot produce an additional solution because because {{{3/2}}} is
positive and a positive number raised to any power, positive or negative,
gives a positive answer. 

Edwin</pre>