Question

729^(2x-5) = 9 Write answer a/b form in simplified form

Answer 1

Given the expression:

$\text{ 729}^{2x\text{ - 5}}\text{ = 9}$

Let's simplify the expression to find x.

$\text{ 729}^{2x\text{ - 5}}\text{ = 9}$$\log _{10}(729^{2x-5})=\log _{10}(9)$$(2x-5_{})\log _{10}(729)=\text{ }\log _{10}(9)$$\text{ 2x - 5 = }\frac{\log_{10}(9)}{\log_{10}(729)}$$\text{ 2x - 5 + 5 = }\log _{729}(9)\text{ + 5 ; Log Rule (}\frac{\log_c\left(b\right)}{\log_c\left(a\right)}=\log _a\mleft(b\mright))$$\text{ 2x = }\log _{3^6}(9)\text{ + 5}$$\text{ 2x = }\frac{1}{6}\log _3(9)\text{ + 5 ; Log Rule }(\log _{a^b}\mleft(x\mright)=\frac{1}{b}\log _a\mleft(x\mright))$$\text{ 2x = }\frac{1}{6}\log _3(3^2)\text{ + 5}$$\text{ 2x = (}\frac{1}{6})(2)\text{ + 5 ; Log Rule (}\log _a\mleft(a^x\mright)=x)$$\text{ 2x = }\frac{2}{6}+5\text{ }\rightarrow\text{ 2x = }\frac{2}{6}\text{ + }\frac{30}{6}$$\text{ 2x = }\frac{32}{6}\text{ }\rightarrow\text{ (2x)(}\frac{1}{2})\text{ = (}\frac{32}{6})(\frac{1}{2})$$\text{ x = }\frac{32}{12}$$\text{ x = }\frac{\frac{32}{4}}{\frac{12}{4}}=\text{ }\frac{8}{3}$

Therefore, x = 8/3.