Exponential equation Exercises

Question

Today I have big problem. Our teacher gave us this HW with exponential equation for marks. I do not want to get bad mark so I am here. Please help me. I did a lot. And I also know how to solve basic exponential equation problems, but not these.

$1)$ $$ 4^{x+1}-768=4^x$$

$2)$ $$\left(\frac{7}{8}\right)^{3x-3}=\left(\frac{64}{49}\right)^{2-x}$$

$3)$ $$ 3^x-1=\frac{72}{3^x}$$

Answer 1

$1) $ Put $y= 4^x$

$2) $ Put $y= \left(\frac{7}{8}\right)^x$

$3) $ Put $y= 3^x$ you will get a quadratic eq. of $y$

Answer 2

For (1) we have: $$4^{x+1}-4^x=768$$ $$4\cdot 4^x-4^x=3\cdot 256$$ $$(4-1)4^x=3\cdot 4^4$$ $$4^x=4^4$$ so $$\color{red}{x=4}.$$

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For (2) instead: $$\left(\frac{7}{8}\right)^{3x-3}=\left[\left(\frac{8}{7}\right)^2\right]^{2-x}$$ $$\left(\frac{7}{8}\right)^{3x-3}=\left(\frac{8}{7}\right)^{2(2-x)}$$ $$\left(\frac{7}{8}\right)^{3x-3}=\left(\frac{7}{8}\right)^{-2(2-x)}$$ $$3x-3=-2(2-x)$$ $$3x-3=-4+2x$$ $$\color{red}{x=-1}.$$

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Finally, (3): multiplying LHS and RHS by $3^x$ we get $${\left(3^x\right)}^2-3^x-72=0.$$ Substituting $y=3^x$ we get $$y^2-y-72=0.$$ $\Delta=289>0$ and $y$ is either $9$ or $-8$. Now $y=3^x=9$ gives $$\color{red}{x=2}$$ while there is no real $x$ such that $y=3^x=-8<0$.