Logarithmic Equations: Find the Value, Keep the Conditions

First, ask what power is missing

Look at $\log_3(2x-1)=2$. The logarithm is asking which power of $3$ produces $2x-1$. Since the answer is $2$, the argument must be $3^2$. You can turn an unfamiliar-looking equation into a familiar linear one.

Before that rewrite, record the domain (परिभाषा-क्षेत्र). A real logarithm $\log_b U$ requires $b>0$, $b\ne1$ and $U>0$. Here the fixed base $3$ already qualifies, while $2x-1>0$ requires $x>\tfrac12$.

The reliable pattern is:

1. State the original restrictions.
2. Use $\log_b U=c\iff U=b^c$.
3. Solve the resulting equation.
4. Substitute the candidate into the original logarithm.

This is an inverse-function rewrite, rather than moving the written base across an equals sign. It works for a fractional base too: $\log_{1/2}U=0$ means $U=(1/2)^0=1$.

Pay attention to what must be positive. The whole argument must be positive; the letter $x$ need not be. For instance, a negative $x$ can make $x+6$ positive. Rejecting every negative answer would lose legitimate solutions.

Try saying the equation aloud before writing a line of algebra: “The base raised to this answer equals this entire argument.” That sentence also helps you keep brackets around expressions such as $2x-1$.

Equal logs point to equal positive arguments

Suppose two readings on the same logarithmic scale are equal. Can the underlying positive numbers be different? No. For any fixed valid base, the logarithm is one-to-one: each output belongs to just one positive input. Therefore

The argument works for both kinds of base. A base above $1$ gives an increasing function; a base between $0$ and $1$ gives a decreasing function. Either function is still one-to-one. Equality does not require an inequality sign reversal.

Write the two argument restrictions separately before simplifying. An equation can appear to have an algebraic answer exactly where one or both original logs fail. For $\log_3(x-1)=\log_3(1-x)$, the first argument requires $x>1$, while the second requires $x<1$. There is no common domain at all. Setting the arguments equal would produce $x=1$, where both arguments are zero; it cannot rescue the equation.

Also check the bases. The direct “equal arguments” step needs the same base on both sides. If the bases differ, first rewrite them using powers or change of base.

This is a useful moment to slow down: identify each complete argument, compare the bases, and only then remove the logarithms through the one-to-one rule. The method becomes quick because you know why each step is allowed.

Combine the logs, then keep the original domain

A sum of logarithms often hides an equation you already know how to solve. With the same valid base and positive arguments, the product law gives $\log_b U+\log_b V=\log_b(UV)$. A difference similarly becomes the logarithm of a quotient.

The important word is positive. The original equation contains two logs, so it needs $U>0$ and $V>0$ individually. After combining, the single expression $\log_b(UV)$ can also exist when both factors are negative. Those new inputs do not belong to the original problem.

For example, $\log_2 x+\log_2(x-2)=3$ begins with $x>2$. Combining gives $\log_2(x(x-2))=3$, then a quadratic. If the quadratic has two real roots, test both against $x>2$; do not assume a quadratic guarantees two logarithmic solutions.

A quotient equation has the same issue. If you begin with $\log_b U-\log_b V$, preserve both original positivity conditions even after writing $\log_b(U/V)$. In contrast, a problem that originally contains only $\log_b(U/V)$ starts with the condition $U/V>0$. The two starting expressions can therefore have different domains.

Use the activity to test a candidate before accepting it. It should pass two checks: the original logarithms must exist, and the two sides must be equal. Defined and solves the equation are different claims.

Name the expression that keeps returning

JEE extension. A repeated logarithm can make an equation look larger than it is. If $\log_2 x$ occurs several times, set $t=\log_2 x$. Solve for $t$, then return to $x$ using $x=2^t$. The substitution changes the notation; it does not remove the original conditions.

For a plain logarithm $t=\log_2 x$ with $x>0$, every real $t$ is possible. If the original expression also contains $\log_4(\log_2 x)$, the inner value must be positive, so now $t>0$. Read nested expressions from the inside out: the output of the inner function becomes the input of the outer function.

Try a simpler pattern first. The equation $(\log_2 x)^2-3\log_2 x+2=0$ becomes $t^2-3t+2=0$. Factoring gives $t=1$ or $t=2$, hence $x=2$ or $x=4$. Both are positive, so both survive.

When bases differ, change them before choosing the repeated expression. Since $4=2^2$, $\log_4 x=\tfrac12\log_2 x$. That makes a nested base-4 expression connect neatly with a base-2 one.

Finish with a check in the full equation, not just in the polynomial for $t$. This habit catches forgotten inner-log restrictions and prevents an answer for the temporary variable from being mistaken for the answer for $x$.

Bring a difficult exponent down

JEE extension. Sometimes the unknown appears in an exponent, and rewriting everything as a single familiar power is awkward. Taking a logarithm of both sides can bring the exponent down: for $A>0$, $\log_b(A^t)=t\log_b A$. Before doing this, check that both sides of the equation are positive.

Consider the positive-domain problem $x^{\log_{10}(x+2)}=x+2$, with $x>0$. Both sides are positive throughout this stated domain. Base 10 is a convenient choice because it is already used in the exponent. After taking logs, the repeated factor $\log_{10}(x+2)$ becomes visible.

Move everything to one side and factor before dividing. If a product is zero, either factor can be zero. Dividing by an expression before checking its zero case may throw away a solution. The simple equation $x^2=x$ illustrates the problem: dividing by $x$ loses the solution 0; factoring $x(x-1)=0$ keeps it.

In a logarithmic equation, a zero-factor candidate must still pass the original domain. Here one factor yields $x=-1$, outside the stated $x>0$ problem. The other gives the valid solution. You are solving the equation on its specified domain, rather than collecting every number produced during rearrangement.

Keep the routine visible: positivity check, logarithm, factor, solve, original check. It works just as well when the chosen logarithm is $\ln$ or another valid fixed base.