Question

I need help on this, please explain how you get the result.

Answer 1

Answer:

$x=5$

Step-by-step explanation:

We have the equation:

$34\cdot 3^{x-2}-2\cdot 3^{x-3}=0.8\cdot 10^{x-2}+10^{x-3}$

Let's simplify this a bit. Let $u=x-2$. Then $u-1=x-3$. We can substitute the exponents:

$34\cdot3^u-2\cdot 3^{u-1}=0.8\cdot 10^u+10^{u-1}$

Use the properties of exponents, we can write $3^{u-1}=\frac{3^u}{3}$. We can do the same thing on the right. So:

$34\cdot3^u-2\cdot \frac{3^u}{3}=0.8\cdot 10^u+\frac{10^u}{10}$

We can now factor out a $3^u$ from the left and a $10^u$ on the right. This yields:

$3^u(34-2\cdot\frac{1}{3})=10^u(0.8+ \frac{1}{10})$

Evaluate the expressions within the parentheses:

$3^u(34-\frac{2}{3})=10^u(\frac{9}{10})$

Evaluate:

$3^u(\frac{100}{3})=10^u(\frac{9}{10})$

Now, let's multiply both sides by $\frac{3}{100}$. So:

$3^u=10^u(\frac{9}{10})(\frac{3}{100})$

Also, let's divide both sides by $10^u$. Multiply on the right:

$\frac{3^u}{10^u}=\frac{27}{1000}$

Therefore:

$3^u=27\text{ and } 10^u=1000$

We can now substitute back u. Notice that 27 is the same as 3 cubed and 1000 is the same as 10 cubed. So:

$3^{x-2}=3^3\text{ and } 10^{x-2}=10^3$

Since they have the same base, their exponents must be equal. Therefore:

$x-2=3$

Add 2 to both sides:

$x=5$

So, the value of x is 5.

And we're done!