Exponents and Logarithm Question Bank -1

Step 1: Express all terms using prime bases

Note that \(6 = 2 \cdot 3\), and \(4 = 2^2\).

So rewrite: \((6^x)(3^{2x+1}) = (2 \cdot 3)^x \cdot 3^{2x+1} = 2^x \cdot 3^x \cdot 3^{2x+1}\)

Now combine powers of 3:   \(2^x \cdot 3^{x + 2x + 1} = 2^x \cdot 3^{3x+1}\)

Right-hand side:   \(4^{x+2} = (2^2)^{x+2} = 2^{2(x+2)} = 2^{2x + 4}\)

So the equation becomes:   \(2^x \cdot 3^{3x+1} = 2^{2x + 4}\)

Step 2: Take natural logarithm of both sides

\(\ln(2^x \cdot 3^{3x+1}) = \ln(2^{2x+4})\)

Apply log laws:   \(\ln(2^x) + \ln(3^{3x+1}) = \ln(2^{2x+4})\)

\(x \ln 2 + (3x+1)\ln 3 = (2x + 4)\ln 2\)

Step 3: Expand and collect like terms

\(x \ln 2 + 3x \ln 3 + \ln 3 = 2x \ln 2 + 4 \ln 2\)

Bring all terms to one side:   \(x \ln 2 – 2x \ln 2 + 3x \ln 3 + \ln 3 – 4 \ln 2 = 0\)

Group like terms:   \(x(-\ln 2 + 3\ln 3) + \ln 3 – 4\ln 2 = 0\)

Step 4: Solve for $x$

\(x(3\ln 3 – \ln 2) = 4 \ln 2 – \ln 3\)

\(x = \frac{4 \ln 2 – \ln 3}{3 \ln 3 – \ln 2}\)

\(x = \frac{\ln(2^4) – \ln(3)}{\ln(3^3) – \ln(2)} = \frac{\ln(16) – \ln(3)}{\ln(27) – \ln(2)} = \frac{\ln\left(\frac{16}{3}\right)}{\ln\left(\frac{27}{2}\right)}\)

Final Answer in the required form: \({x = \frac{\ln\left(\frac{16}{3}\right)}{\ln\left(\frac{27}{2}\right)}}\)

Here, \(a = \frac{16}{3}\) and \(b = \frac{27}{2}\), both rational.

Step 1: Express all terms using the same base

Note that \(9 = 3^2\), so rewrite:

\((3^2)^x + 4(3^x) – 12 = 0\)

\(3^{2x} + 4(3^x) – 12 = 0\)

Step 2: Make a substitution

Let \(y = 3^x\). Then \(3^{2x} = (3^x)^2 = y^2\).

Now the equation becomes:   \(y^2 + 4y – 12 = 0\)

Step 3: Solve the quadratic equation

Use the quadratic formula:

\(y = \frac{-4 \pm \sqrt{(4)^2 + 4 \cdot 1 \cdot 12}}{2 \cdot 1}\)

\(y = \frac{-4 \pm \sqrt{16 + 48}}{2}\)

\(y = \frac{-4 \pm \sqrt{64}}{2}\)

\(y = \frac{-4 \pm 8}{2}\)

So, \(y = 2\) or \(y = -6\)

Step 4: Back-substitute \(y = 3^x\)

1. \(3^x = 2 \Rightarrow x = \log_3 2\)
2. \(3^x = -6\) has no solution since \(3^x > 0\) for all real \(x\)

Final Answer:   \(x = \log_3 2\)

Step 1: Convert logarithmic form to exponential form

Recall that: \(\log_b A = C \Rightarrow A = b^C\)

Apply this rule:   \(\sqrt[3]{100 – x^2} = 16^{1/2}\)

Step 2: Simplify right-hand side

\(16^{1/2} = \sqrt{16} = 4\)

So,   \(\sqrt[3]{100 – x^2} = 4\)

Step 3: Eliminate the cube root

Cube both sides:    \(100 – x^2 = 4^3 = 64\)

Step 4: Solve for \(x\)

\(x^2 = 100 – 64 = 36\)

\(x = \pm 6\)

Final Answer:   \(x = 6\) or \(x = -6\)

Step 1: Express both sides with the same base

We know:  \(\sqrt{3} = 3^{1/2}\)

So the right-hand side becomes:  \((\sqrt{3})^{126} = (3^{1/2})^{126} = 3^{63}\)

Step 2: Equate the exponents

Since both sides have the same base (3), equate the powers:  \(x^2 – 1 = 63\)

Step 3: Solve for \(x\)

\(x^2 = 63 + 1 = 64\)

\(x = \pm 8\)

Final Answer:  \(x = 8\) or \(x = -8\)

Step 1: Express all powers with the same base

Note that:

\(4^x = (2^2)^x = 2^{2x}\)

\(8^y = (2^3)^y = 2^{3y}\)

So the second equation becomes:   \(2^{2x} = 2^{3y} \Rightarrow 2x = 3y\)

Step 2: Substitute into the linear equation

From \(2x = 3y\), we get:   \(x = \frac{3}{2}y\)

Now substitute into the first equation:   \(\frac{3}{2}y + 2y = 5\)

Convert to a single fraction:   \(\frac{3y + 4y}{2} = 5 \Rightarrow \frac{7y}{2} = 5\)

Multiply both sides by 2:   \(7y = 10 \Rightarrow y = \frac{10}{7}\)

Step 3: Find the value of \(x\)

Substitute \(y = \frac{10}{7}\) into \(x = \frac{3}{2}y\):

\(x = \frac{3}{2} \cdot \frac{10}{7} = \frac{30}{14} = \frac{15}{7}\)

Final Answer  \({x = \frac{15}{7}, \quad y = \frac{10}{7}}\)

Step 1: Use the definition of logarithm

From \(\log_x y = 1\), rewrite it in exponential form:

\(x^1 = y\)

So we have:   \(y = x\)

Step 2: Substitute into the second equation

Substitute \(y = x\) into \(xy = 16\):

\(x \cdot x = 16\)

\(x^2 = 16\)

Step 3: Solve for \(x\)

\(x = \sqrt{16} = 4\) (since \(x > 0\))

Now substitute back to find \(y\):

\(y = x = 4\)

Final Answer \({x = 4,\quad y = 4}\)

Identify the domain condition

The expression inside the logarithm must be strictly positive:

\(4x^2 – rx + r – 1 > 0\) for all \(x \in \mathbb{R}\).

This is a quadratic inequality. A quadratic is always positive if:

• Its leading coefficient is positive (which is true here, since 4 > 0), and

• Its discriminant is less than 0 (no real roots, hence the expression never equals or crosses 0).

Apply the discriminant condition

The discriminant of \(4x^2 – rx + (r – 1)\) is:  \(\Delta = (-r)^2 – 4(4)(r – 1) = r^2 – 16r + 16\)

We require: \(r^2 – 16r + 16 < 0\)

Solve the inequality

First solve the equation: \(r^2 – 16r + 16 = 0\)

Use the quadratic formula:

\(r = \frac{16 \pm \sqrt{(-16)^2 – 4(1)(16)}}{2}\)

\(\quad = \frac{16 \pm \sqrt{256 – 64}}{2}\)

\(\quad = \frac{16 \pm \sqrt{192}}{2}\)

\(\quad = \frac{16 \pm 8\sqrt{3}}{2}\)

\(\quad = 8 \pm 4\sqrt{3}\)

Write the final condition

For the discriminant to be negative,  \({8 – 4\sqrt{3} < r < 8 + 4\sqrt{3}}\)

Final Answer. \({8 – 4\sqrt{3} < r < 8 + 4\sqrt{3}}\)

This ensures that \(h(x)\) is defined for all real values of \(x\).

Use logarithmic rules on the first equation

\(\log_2 (6x) = 1 + 2\log_2 y\)

Since \(1 = \log_2 2\), rewrite the RHS:

\(\log_2 (6x) = \log_2 2 + 2\log_2 y\)

Now apply the power rule and addition rule:

\(\log_2 (6x) = \log_2 2 + \log_2 y^2 = \log_2 (2y^2)\)

So: \(6x = 2y^2 \quad \text{(Equation 1)}\)

Use logarithmic rules on the second equation

\(1 + \log_6 x = \log_6 (15y – 25)\)

Again, use \(1 = \log_6 6\):

\(\log_6 6 + \log_6 x = \log_6 (15y – 25)\)

Apply the addition rule: \(\log_6 (6x) = \log_6 (15y – 25)\)

So: \(6x = 15y – 25 \quad \text{(Equation 2)}\)

Substitute Equation 1 into Equation 2

From Equation 1: \(6x = 2y^2\)

Substitute into Equation 2: \(2y^2 = 15y – 25\)

Rearrange: \(2y^2 – 15y + 25 = 0\)

Factor: \((2y – 5)(y – 5) = 0\)

So: \(y = \frac{5}{2} \quad \text{or} \quad y = 5\)

Find corresponding values of x

Use Equation 1: \(6x = 2y^2 \Rightarrow x = \frac{2y^2}{6} = \frac{y^2}{3}\)

If \(y = \frac{5}{2}\):   \(x = \frac{(5/2)^2}{3} = \frac{25}{12}\)

If \(y = 5\):   \(x = \frac{25}{3}\)

Final Answers. \({(x = \frac{25}{12},\ y = \frac{5}{2}) \quad \text{or} \quad (x = \frac{25}{3},\ y = 5)}\)

Convert both logarithms to the same base

We choose base 4. Use the change of base rule:

\(\log_{16} (13 – 4x) = \frac{\log_4 (13 – 4x)}{\log_4 16}\)

Now, recall:  \(\log_4 16 = 2\), since \(16 = 4^2\)

So the equation becomes:  \(\log_4 (2 – x) = \frac{\log_4 (13 – 4x)}{2}\)

Eliminate the fraction by multiplying both sides by 2

\(2 \log_4 (2 – x) = \log_4 (13 – 4x)\)

Use the logarithmic identity:  \(r \log_b A = \log_b (A^r)\)

So:  \(\log_4 (2 – x)^2 = \log_4 (13 – 4x)\)

Remove logarithms (equal bases)

\((2 – x)^2 = 13 – 4x\)

Expand the LHS:  \(4 – 4x + x^2 = 13 – 4x\)

Subtract \(13 – 4x\) from both sides:  \(x^2 – 9 = 0\)

Solve the quadratic

\(x^2 = 9 \Rightarrow x = \pm 3\)

Now check domain validity:

• For \(x = 3\):  \(2 – x = -1\) → Not valid for log input
• For \(x = -3\):  \(2 – (-3) = 5\), and \(13 – 4(-3) = 25\), both positive

Final Answer.  \({x = -3}\)

Step 1: Rearrange the inequality

Move all terms to one side:  \((\ln x)^2 – (\ln 2)(\ln x) – 2(\ln 2)^2 < 0\)

Step 2: Let \(y = \ln x\) to simplify

Substitute and rewrite as a quadratic:  \(y^2 – (\ln 2)y – 2(\ln 2)^2 < 0\)

Step 3: Solve the corresponding quadratic equation

\(y^2 – (\ln 2)y – 2(\ln 2)^2 = 0\)

Factor:  \((y – 2\ln 2)(y + \ln 2) = 0\)

So the roots are:  \(y = 2\ln 2\) or \(y = -\ln 2\)

Step 4: Solve the inequality

We now solve:  \((y – 2\ln 2)(y + \ln 2) < 0\)

This quadratic is negative between its roots:  \(-\ln 2 < y < 2\ln 2\)

Step 5: Substitute back \(y = \ln x\)

\(-\ln 2 < \ln x < 2\ln 2\)

Apply exponential to all parts:  \(e^{-\ln 2} < x < e^{2\ln 2}\)

Use log identities:  \(\frac{1}{2} < x < 4\)

Final Answer.  \({\frac{1}{2} < x < 4}\)