The value of ${\log }_a\left(\frac{a}{b}\right)+{\log }_b\left(\frac{b}{a}\right)$ for $a>1$ and $b>1$ cannot be equal to

The correct answer is 1. The expression ${\log }_a\left(\frac{a}{b}\right)+{\log }_b\left(\frac{b}{a}\right)$ can never equal 1. Here is a full, step-by-step derivation of why.

Step 1 — Expand each logarithm using the Quotient Rule

Recall that ${\log }_x\left(\frac{p}{q}\right)={\log }_x\left(p\right)-{\log }_x\left(q\right)$.

Applying this to the first term:

${\log }_a\left(\frac{a}{b}\right)={\log }_a\left(a\right)-{\log }_a\left(b\right)=1-{\log }_a\left(b\right)$

Applying this to the second term:

${\log }_b\left(\frac{b}{a}\right)={\log }_b\left(b\right)-{\log }_b\left(a\right)=1-{\log }_b\left(a\right)$

Step 2 — Add the two terms

The full expression becomes:

$(1-{\log }_a(b\left)\right)+(1-{\log }_b(a\left)\right)=2-{\log }_a\left(b\right)-{\log }_b\left(a\right)$

Step 3 — Introduce a substitution

Let $t={\log }_a\left(b\right)$. Since $a>1$ and $b>1$, we know that $t>0$.

By the change-of-base reciprocal identity:

${\log }_b\left(a\right)=\frac{1}{t}$

So the expression simplifies to the function:

$f\left(t\right)=2-t-\frac{1}{t},t>0$

Step 4 — Find the maximum of f(t) using AM-GM Inequality

The AM-GM Inequality states that for any two positive real numbers $p$ and $q$:

$\frac{p+q}{2}\ge \sqrt{pq}$

Applying this with $p=t$ and $q=\frac{1}{t}$ (both positive for $t>0$):

$\frac{t+\frac{1}{t}}{2}\ge \sqrt{t\cdot \frac{1}{t}}=\sqrt{1}=1$

Therefore:

$t+\frac{1}{t}\ge 2\text{for all}t>0$

Multiplying both sides by $-1$ (which flips the inequality):

$-t-\frac{1}{t}\le -2$

Adding 2 to both sides:

$2-t-\frac{1}{t}\le 0$

That is:

$f\left(t\right)\le 0\text{for all}t>0$

Step 5 — Interpret the result

The expression is always less than or equal to 0 when $a>1$ and $b>1$. The maximum value of 0 is achieved only when $t=1$, i.e., when $a=b$.

Now let's check the given options against this finding:

• 0: Achievable when $a=b$. ✓
• −0.5: A negative value, so achievable. ✓
• −1: A negative value, so achievable. ✓
• 1: This is positive, which is strictly greater than the maximum possible value of 0. ✗ — impossible!

Conclusion: Since the expression ${\log }_a\left(\frac{a}{b}\right)+{\log }_b\left(\frac{b}{a}\right)\le 0$ for all valid $a,b>1$, it can never equal 1. The correct answer is 1.