An algebra problem by Mayur Ranaware

Write $\frac{xy}{x+y}=\frac{1}{\dfrac{x+y}{xy}}=\frac{1}{\dfrac{1}{x}+\dfrac{1}{y}}$ From $5^x=30^7$ , we get $x=7\log_5 {30}$ . And from $6^y=30^7$ , we get $y=7 \log_6 {30}$ . Thus, $\frac{1}{\dfrac{1}{x}+\dfrac{1}{y}}=\frac{1}{\dfrac{1}{7\log_5 {30}}+\dfrac{1}{7\log_6 {30}}}=\frac{1}{\dfrac{1}{7} \log_{30} 5+\dfrac{1}{7} \log_{30} 6}=\frac{1}{\dfrac{1}{7} \log_{30} {30}}=\boxed{7}.$

Let

${5}^{x} = {6}^{y} = {30}^{7} = k$

Then, ${5}^{x} = k$ => $5 = {k}^{\frac{1}{x}}$

Similarly, $6 = {k}^{\frac{1}{y}}$

Also, 5 * 6 = 30

Therefore, ${k}^{\frac{1}{x}}$ * ${k}^{\frac{1}{y}}$ = ${k}^{\frac{x+y}{xy}}$ = 30

We know that k = ${30}^{7}$

Hence, $30 = {30}^{7(\frac{x + y}{xy})}$

therefore, $\frac{xy}{x+y} = 7$

We know that $5^x \times 6^y = (30^7)^2 = 30^{14} = (5^{14} \times 6^{14})$ Therefore, $x = y = 14$ Hence, we have: $\frac{14 \cdot 14}{14 + 14} = 7$

1) 6^y = 30^7 ===> y* ln6 = 7 * ln30

2) 5^x = 30^7 ===> x* ln5 = 7* ln30

so, y * ln6 = x * ln 5 ====> y = x * ln5 / ln6

substituting

x * y = x^2 * ln5/ln6 and

x + y = x + x * ln5/6 ====> x * (1 + ln5/ln6) ===> x * (ln6 + ln5)/ln6

To find value of expression, substitute:

(x^2 ln5/ln6) / (x * (ln6 + ln5)/ln6 ====> x ln5/ln6 * ln6/(ln6 + ln5)

= x ln5 / (ln6 + ln5) = x * ln5 / ln(6 5) = x ln5 / ln30

Substitute from 2):

= 7 * ln30 / ln30 = 7

5^x = 30^7 => 5^x = (5^7)(6^7) => 5^(x-7) = 6^7. Taking log on both sides, (x-7)log5 = 7log6 => log5/log6 = 7/(x-7) --- (i). Also, 6^y = 30^7 => 6^(y-7) = 5^7. Taking log on both sides, (y-7)log6 = 7log5 => log5/log6 = (y-7)/7 --- (ii). Comparing equation (i) & (ii), we have 7/(x-7) = (y-7)/7. Simplifying this equation and we get xy/(x+y) = 7.

30^7=(5.6)^7 =5^7.6^7 Squaring both sides we get, 5^7.6^7.5^7.6^7=5^x.5^x so, 5^7.6^7.5^7.6^7=5^x.6^y therefore, x=y=14 rest is easy

$5^x=6^y=30^7=5^7\cdot 6^7$

$log_55^{x-7}= log_5 6^7$

$x-7=log_5 279936$

$x=14.794$

similarly $y=13.288$

Hence the required value is $6.821i.e.\boxed{7}$