An algebra problem by Paola Ramírez

$\large 7^{x+7}=7^x*7^7=8^x~~\therefore ~7^7=\left (\dfrac{8}{7} \right )^x \\ \large x=\log_{b} 7^7=\log_{b}\left (\dfrac{8}{7} \right )^x= x*\log_{b}\left (\dfrac{8}{7} \right )\\ \large \implies~1= \log_{b}\left (\dfrac{8}{7} \right )\\ \large \implies~~b^1= \dfrac{8}{7} \\ \large m+n= \boxed{\color{#D61F06}{15}}$

$7^{x+7}=8^{x}$

$\log 7^{x+7}=\log 8^x$

$(x+7)\log 7=x\log 8$

$x(\log 8-log 7)=\log 7^{7}$

$x= \frac{ \log 7^{7} } {\log 8-\log 7}= \frac{ \log 7^{7} } {\log \frac{8}{7}}$

Apply change of base:

$x=\frac{\frac{ \log_{\frac{8}{7}} 7^7}{ \log_{\frac{8}{7}} 10}}{\frac{ \log_{\frac{8}{7}} \frac{8}{7}}{ \log_{\frac{8}{7}} 10}}=\log_{\frac{8}{7}} 7^7$

$\therefore b=\frac{8}{7}$ and $\boxed{m+n=15}$

(7^7)(7^x) = 8^x

7^7 = (8/7)^x

Taking the logarithm for the base b

log (7^7) = x log (8/7) = x

Then

log (8/7) = 1

Then

b = 8/7 = m/n

Then

m + n = 8 + 7 = 15

7^(x + 7) = 8^x (x + 7)ln(7) = xln(8) x = 7ln(7)/(ln(8/7)) = 7*log(base 8/7) 7 Therefore b = 8/7 m + n = 15 OR

Simply from 1st equation, we get

7^7 = (8/7)^x

Substituting the value of 7^7 in 2nd equation, we get

x = log b (8/7)^x where b is the base of the log

Thus,

x = x log b (8/7)

Thus, we get

b = 8/7 = m/n

Therefore, m = 8 and n = 7

So, m+n = 15

Since, $x= \log _{ b }{ 7^{ 7 } }$ , therefore, $b^{ x }=7^{ 7 } \rightarrow (1)$ , it is also apparent that $b > 0$ .

Then,

Since, $7^{x+7} = 8^x$ , therefore $7^x \cdot 7^7 = 8^x$ , then substituting from $(1)$ , therefore, $7^x \cdot b^x = 8^x$ .

Now taking $x^{th}$ root of both sides ,

Therefore, $\sqrt[x]{7^x \cdot b^x} = \sqrt[x]{8^x}$ ,

Then , $7 \cdot b = 8$ ,

Therefore, $b = \frac{8}{7} = \frac{m}{n}$ ,

Therefore, $m+n = 8+7 = \boxed{15}$ .