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Let $p = 2x + 3$ and $q = 3x + 7$ . Then the given equation can be written as

$\log_{p}(p * q) = 4 - \log_{q}(p^{2})$ $\Longrightarrow$ $1 + \log_{p}(q) = 4 - 2 * \log_{q}(p)$ .

Now by the change-of-base rule we have that $\log_{p}(q) = \dfrac{1}{\log_{q}(p)}$ .

So letting $z = \log_{q}(p)$ , and assuming that $z \ne{0}$ , i.e., that $p \ne{1}$ , our equation becomes

$\dfrac{1}{z} = 3 - 2z \Longrightarrow 2z^{2} - 3z + 1 = 0$ ,

which has solutions $z = 1$ and $z = \frac{1}{2}$ . Now with $z = 1$ we have that

$\log_{q}(p) = 1 \Longrightarrow q = p \Longrightarrow x = -4$ ,

which would give us negative bases, and hence we can discard this value of $z$ . With $z = \frac{1}{2}$ , we have that

$\log_{q}(p) = \frac{1}{2} \Longrightarrow p = \sqrt{q}$ .

After substituting for $p$ and $q$ , squaring both sides and simplifying, we end up with $4x^{2} + 9x + 2 = 0$ , which has solutions $x = -2$ and $x = -\frac{1}{4}$ . The first of these makes $p \lt 0$ , so we are left with just $x = -\frac{1}{4}$ , which does check out when plugging it back into the original equation.

Thus we have $a = 1, b = 4$ , giving us the solution $a + b = \boxed{5}$ .

We rearrange the equation, factorizing and using the notation

${ Log }_{ a }(x)=Log[a,x]$ , and then do the following steps

$Log[2x+3,(3x+7)(2x+3)]-3=1-Log[3x+7,{ (2x+3) }^{ 2 }]$

$Log[2x+3,(3x+7)(2x+3)]-Log[2x+3,{ (2x+3) }^{ 3 }]=$

$Log[3x+7,3x+7]-Log[3x+7,{ (2x+3) }^{ 2 }]$

$Log[2x+3,\dfrac { 3x+7 }{ { (2x+3) }^{ 2 } } ]=Log[3x+7,\dfrac { 3x+7 }{ { (2x+3) }^{ 2 } } ]$

which means that

$\dfrac { 3x+7 }{ { (2x+3) }^{ 2 } } =1$

which leads to a quadratic equation with two roots $x=-2,-\dfrac { 1 }{ 4 }$ , but $x=-2$ results in a negative base and delivers indeterminate results, leaving $x=-\dfrac { 1 }{ 4 } =-\dfrac { a }{ b }$ as the only possible root, and thus $a+b=5$

Get the factors of each trinomial, you will get log(3x+7)(2x+3) base (2x+3) = 4 - 2log(2x+3). Then apply properties of logarithms, you'll arrive at:

log (3x+7) base (2x+3) +1 = 4-2log(2x+3) base (3x+7)

Apply another property of logarithm, which is the change of base, the equation becomes:

log(3x+7)/log (2x+3) + 2log(2x+3)/log(3x+7) = 3

Recall: 2log(2x+3)/log(3x+7) = 1/ [2log(2x+3)/log(3x+7)]

this becomes log(3x+7)/2log (2x+3)

log(3x+7)/log (2x+3) + log(3x+7)/2log (2x+3) = 3

Combine the two fractions; this becomes

3log(3x+7)/2log (2x+3) = 3

3 is cancelled and 2log (2x+3) is transposed to the other side of the equation.

log(3x+7) = 2log (2x+3)

log is cancelled and the two becomes the exponent of (2x+3)

3x+7 = 4x^2+12x+9

Combine like terms and rearrange:

4x^2+9x+2=0 <---- get the factors of this trinomial, this decomposes into:

(4x+2)(x+1)=0

Values of x are -1/4 and -1, but -1 is an extraneous root. So -1/4 is the answer and the sum of its digits is 1+4 = 5