Q.With the circuit diagram show how to obtain an output which is logarithm of input.

A logarithmic amplifier gives an output proportional to the natural logarithm of the input; it uses an op-amp with a diode in the feedback path, exploiting the diode's exponential current-voltage law. …

[!TLDR]

A diode in the feedback path of an op-amp converts its exponential I-V law into a logarithm: with a virtual ground at the input, $V_O = -\eta V_T\,\log_e\!\left(\dfrac{V_i}{I_s R_i}\right)$, so $V_O \propto \log_e(V_i)$.

Circuit: The input $V_i$ is applied through $R_i$ to the inverting terminal (node A) of the op-amp; a semiconductor diode is connected in the feedback path between the output and node A; the non-inverting terminal (node B) is grounded.

Derivation: For an ideal op-amp, open-loop gain $A=\infty$ and input impedance $Z_i=\infty$, so $V_B = V_A = 0$ (virtual ground) and $I_B = 0$.

Applying KCL at node A:

Since $V_A = 0$:

The voltage across the diode is

From Shockley's diode equation,

Substituting (1) and (2) in (3):

Taking $\log_e$ on both sides:

$\boxed{V_O = -\eta V_T\,\log_e\!\left(\frac{V_i}{I_s R_i}\right)}$ …