if logx to the base a,logx to the base b,logx to the base c are in A.P. where x is not equal to 1,then show that c^2=(ac)^logb to the base a

To tackle the problem where the logarithms of \( x \) to different bases \( a \), \( b \), and \( c \) are in arithmetic progression (A.P.), we need to start by recalling the definition of A.P. For three numbers to be in A.P., the middle number must be the average of the other two. In this case, we have \( \log_a x \), \( \log_b x \), and \( \log_c x \) in A.P.

Setting Up the Equation

Given that \( \log_a x \), \( \log_b x \), and \( \log_c x \) are in A.P., we can express this relationship mathematically:

\[ \log_b x = \frac{\log_a x + \log_c x}{2} \]

Using Change of Base Formula

We can apply the change of base formula for logarithms, which states that:

\[ \log_b x = \frac{\log_a x}{\log_a b}, \quad \log_c x = \frac{\log_a x}{\log_a c} \]

Substituting these into our A.P. equation gives us:

\[ \frac{\log_a x}{\log_a b} = \frac{\log_a x + \frac{\log_a x}{\log_a c}}{2} \]

Clearing the Denominators

To simplify this equation, we can multiply through by \( 2 \log_a b \log_a c \) to eliminate the fractions:

\[ 2 \log_a c \cdot \log_a x = \log_a b \cdot \log_a x + \log_a x \cdot \log_a b \]

Factoring Out Common Terms

Now, we can factor out \( \log_a x \) from the right side:

\[ 2 \log_a c \cdot \log_a x = \log_a x \left( \log_a b + \log_a c \right) \]

Assuming \( \log_a x \neq 0 \) (which is valid since \( x \neq 1 \)), we can divide both sides by \( \log_a x \):

\[ 2 \log_a c = \log_a b + \log_a c \]

Rearranging the Equation

Rearranging gives us:

\[ \log_a c = \log_a b \]

Exponentiating to Eliminate Logarithms

Exponentiating both sides results in:

\[ c = b \]

Final Steps to Prove the Given Equation

Now, we need to show that \( c^2 = (ac)^{\log_b a} \). Since we have established that \( c = b \), we can substitute \( b \) for \( c \) in the equation:

\[ b^2 = (ab)^{\log_b a} \]

Using properties of exponents, we can rewrite the right side:

\[ b^2 = a^{\log_b a} \cdot b^{\log_b a} \]

Since \( a^{\log_b a} = b \) (by the definition of logarithms), we can simplify further:

\[ b^2 = b^{\log_b a} \cdot b^{\log_b a} = b^2 \]

This confirms that both sides are equal, thus proving the statement. Therefore, we have shown that:

\[ c^2 = (ac)^{\log_b a} \]

Summary

In summary, we started with the condition of logarithms being in A.P., applied the change of base formula, manipulated the resulting equations, and ultimately proved the required relationship. This process highlights the interconnectedness of logarithmic properties and arithmetic sequences.