Can't solve this exponential equation: $5^{x+1}-3\cdot 5^{x-1} - 6\cdot 5^x+10 = 0$

Question

How does one solve for $x$ in the following:

$$5^{x+1}-3\cdot 5^{x-1} - 6\cdot 5^x+10 = 0$$

Answer 1

write it in the form $$5\cdot 5^x-\frac{3}{5}\cdot 5^x-6\cdot 5^x+10=0$$ setting $5^x=t$ so you will get a linear equation

Answer 2

Dividing both sides by $5^{x-1}$ (this is valid since for all real numbers, $5^{x-1}$ is nonzero) we are left with the following:

$\frac{5^{x+1}}{5^{x-1}}-3\frac{5^{x-1}}{5^{x-1}}-6\frac{5^{x}}{5^{x-1}}+10\frac{1}{5^{x-1}}=25-3-30+10\cdot 5^{1-x}=0$

Yielding the equation:

$10\cdot 5^{1-x}=8$

You can continue solving from there.

Answer 3

First assume $5^x=y$ solving that equation you get $y=5^x=\frac{25}{4}$ so now $5^x=25/4$ taking log to the base $5$ you get $x=2-log_{5}(4)$ so now $log_{5}4=log_{10}4/log_{10}5\approx 0.85$ thus $x=2-0.85=1.15$ . And its done.

Answer 4

$5^{x+1}-3*5^{x-1}-6*5^x=-10$

Putting $5^x=y$,

$5y-\frac{3y}{5}-6y=-10$

Solving,we get $y=\frac{25}{4}$

Answer 5

Notice, $$5^{x+1}-3\cdot 5^{x-1}-6\cdot 5^x+10=0$$ $$6\cdot 5^x+3\cdot 5^{x-1}-5^{x+1}=10$$

$$5^{x-1}(6\cdot 5+3-5^2)=10$$ $$8\cdot 5^{x-1}=10$$ $$5^{x-1}=\frac{5}{4}$$ $$(x-1)\ln(5)=\ln\left(\frac{5}{4}\right)$$ $$x=\frac{\ln\left(\frac{5}{4}\right)}{\ln(5)}+1=\frac{\ln\left(\frac{5}{4}\right)+\ln(5)}{\ln(5)}$$

$$\bbox[5px, border:2px solid #C0A000]{\color{blue}{x=\frac{\ln\left(\frac{25}{4}\right)}{\ln(5)}}}$$