Find the value of $x$.

Question

$$7^{x-2}-4 \cdot 7^{x-3} = 147$$

Find the value of $x$

Let's call $7^{x-2} = y$

$$y-4\cdot \frac{1}{2}y = 147$$

However, there's no solution root as far as I can determine.

Regards

Answer 1

To solve this problem you need to know that any number $y$ (not the one from your substitution) can be written as:
$$y^{a}=y^{a-b+b}=y^{a+b}\cdot y^{-b}=y^{a+b}\cdot \frac{1}{y^b}$$
In your example:
$$7^{x-3}=7^{x-2-1}=7^{x-2}\cdot 7^{-1}=7^{x-2}\cdot \frac{1}{7^1}$$

Answer 2

With $y=7^{x-2}$ you have $$7^{x-3}=7^{x-2-1}=\frac{1}{7} 7^{x-2}$$ so$$y-4 \cdot \frac{1}{7} y=147$$ which has a solution $>0$ as $1-\frac{4}{7} >0$.

Answer 3

$$ u= 7^{x-3} \implies 7u-4u=147 \implies u=49 \implies 7^{x-3}=7^2 \implies x=5$$

Answer 4

$7^{x-2}(1-\frac {4}{7})=3\times 7^2$

$7^{x-3}=7^2$

$x-3=2$

$x=5$