Answer :

Sure! Let's solve the given equation step by step. We need to solve the equation:

[tex]\[ \left(\sqrt{\frac{3}{5}}\right)^{x-1} = \left(\frac{27}{125}\right)^{-1} \][/tex]

### Step 1: Simplify the right side of the equation

First, we can simplify the right-hand side of the equation. Recall that:

[tex]\[ \left(\frac{a}{b}\right)^{-1} = \frac{b}{a} \][/tex]

So:

[tex]\[ \left(\frac{27}{125}\right)^{-1} = \frac{125}{27} \][/tex]

### Step 2: Substitute and rewrite the equation

Now the equation becomes:

[tex]\[ \left(\sqrt{\frac{3}{5}}\right)^{x-1} = \frac{125}{27} \][/tex]

### Step 3: Simplify the left side of the equation

Next, look at the left-hand side of the equation. Recall that:

[tex]\[ \sqrt{\frac{3}{5}} = \left(\frac{3}{5}\right)^{1/2} \][/tex]

So we rewrite the left-hand side as:

[tex]\[ \left(\left(\frac{3}{5}\right)^{1/2}\right)^{x-1} \][/tex]

Using the power of a power rule [tex]\((a^m)^n = a^{mn}\)[/tex], we get:

[tex]\[ \left(\frac{3}{5}\right)^{(1/2)(x-1)} = \left(\frac{3}{5}\right)^{\frac{x-1}{2}} \][/tex]

### Step 4: Equate and solve exponents

We now have:

[tex]\[ \left(\frac{3}{5}\right)^{\frac{x-1}{2}} = \frac{125}{27} \][/tex]

Notice that [tex]\(\frac{125}{27}\)[/tex] can be written as [tex]\(\left(\frac{5}{3}\right)^3\)[/tex]:

[tex]\[ \frac{125}{27} = \left(\frac{5}{3}\right)^3 \][/tex]

We then rewrite the equation as:

[tex]\[ \left(\frac{3}{5}\right)^{\frac{x-1}{2}} = \left(\frac{5}{3}\right)^3 \][/tex]

### Step 5: Align bases and solve for x

Since [tex]\(\frac{3}{5}\)[/tex] is the reciprocal of [tex]\(\frac{5}{3}\)[/tex], we can rewrite [tex]\(\left(\frac{5}{3}\right)^3\)[/tex] as [tex]\(\left(\frac{3}{5}\right)^{-3}\)[/tex]:

[tex]\[ \left(\frac{3}{5}\right)^{\frac{x-1}{2}} = \left(\frac{3}{5}\right)^{-3} \][/tex]

If the bases are equal, we can set their exponents equal to each other:

[tex]\[ \frac{x-1}{2} = -3 \][/tex]

### Step 6: Solve for x

Multiply both sides by 2 to clear the fraction:

[tex]\[ x - 1 = -6 \][/tex]

Finally, add 1 to both sides to isolate x:

[tex]\[ x = -6 + 1 \][/tex]
[tex]\[ x = -5 \][/tex]

So, the solution to the equation is [tex]\( x = -5 \)[/tex].