# I got a problem in indices [closed]

$$3^{(2x+3)} - 2.9^{(x+1)} =1/3$$

• Do you mean $2.9$ or $2\times9$? Jul 16, 2020 at 10:13
• I am tempted to say it is $2\times9^{(x+1)}$, otherwise the solutions would be way more difficult, as logarithms would be involved. Jul 16, 2020 at 10:47
Using the index laws, transform $$3^{2x+3} = 3^{2x} \times 3^3$$ and $$9^{x+1} = 3^{2(x+1)} = 3^{2x}\times 3^2$$, so that the LHS becomes $$3^{2x}(3^3 - 2\times 3^2) = 3^{2x}(3\times 3^2 - 2\times 3^2) = 3^{2x} \times 3^2$$. Now you need only to solve $$3^{2x} \times 3^2 = 3^{-1}$$ which means $$2x + 2 = -1$$, so $$x = -3/2$$.