Solve the questions and the answers should be in English

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June 2023
Given: $m \log_3 2 = 10 \log_3 6$ . Find the value of $m$ .

Apply the power rule of logarithms, $a \log_b c = \log_b c^a$ , to both sides of the equation.
$\log_3 2^m = \log_3 6^{10}$
Since the logarithms have the same base and are equal, their arguments must be equal.
$2^m = 6^{10}$
Take the common logarithm (base 10) on both sides.
$\log(2^m) = \log(6^{10})$
Apply the power rule of logarithms again.
$m \log 2 = 10 \log 6$
Solve for $m$ .
$m = \frac{10 \log 6}{\log 2}$
This can also be expressed using the change of base formula $\frac{\log_b A}{\log_b C} = \log_C A$ .
$m = 10 \log_2 6$ $m = 10 \log_2 (2 \times 3)$ $m = 10 (\log_2 2 + \log_2 3)$ $m = 10 (1 + \log_2 3)$
The exact value is:
$m = 10(1 + \log_2 3)$
The numerical approximation is:
$m \approx 10(1 + 1.58496)$ $m \approx 10(2.58496)$ $m \approx 25.85$
The question asks for "the value of m", so an exact logarithmic expression is appropriate.

The value of $m$ is $\boxed{10(1 + \log_2 3)}$ .

June 2018
Find the value of $x$ for which $y = 2 \log_2 x$ and $y + 4 = \log_2 (2x)$ .
Substitute the first equation ( $y = 2 \log_2 x$ ) into the second equation.
$2 \log_2 x + 4 = \log_2 (2x)$
Use the product rule of logarithms, $\log_b (AB) = \log_b A + \log_b B$ , on the right side.
$2 \log_2 x + 4 = \log_2 2 + \log_2 x$
Since $\log_2 2 = 1$ :
$2 \log_2 x + 4 = 1 + \log_2 x$
Rearrange the equation to isolate $\log_2 x$ .
$2 \log_2 x - \log_2 x = 1 - 4$ $\log_2 x = -3$
Convert the logarithmic equation to an exponential equation using the definition $\log_b A = C \implies A = b^C$ .
$x = 2^{-3}$
Simplify the expression for $x$ .
$x = \frac{1}{2^3}$ $x = \frac{1}{8}$
The value of $x$ is $\boxed{\frac{1}{8}}$ .

June 2017
Given that $2^x = 3^y$ and that $x+y=1$ , show that $x = \frac{\log 3}{\log 6}$ .
From the equation $x+y=1$ , express $y$ in terms of $x$ .
$y = 1-x$
Substitute this expression for $y$ into the equation $2^x = 3^y$ .
$2^x = 3^{1-x}$
Take the common logarithm (base 10, denoted as $\log$ ) on both sides of the equation.
$\log(2^x) = \log(3^{1-x})$
Apply the power rule of logarithms, $\log A^B = B \log A$ , to both sides.
$x \log 2 = (1-x) \log 3$
Distribute $\log 3$ on the right side.
$x \log 2 = \log 3 - x \log 3$
Gather all terms containing $x$ on one side of the equation.
$x \log 2 + x \log 3 = \log 3$
Factor out $x$ from the terms on the left side.
$x (\log 2 + \log 3) = \log 3$
Use the product rule of logarithms, $\log A + \log B = \log (AB)$ , on the expression in the parenthesis.
$x \log (2 \times 3) = \log 3$ $x \log 6 = \log 3$
Solve for $x$ .
$x = \frac{\log 3}{\log 6}$
This matches the required expression.

We have shown that $x = \boxed{\frac{\log 3}{\log 6}}$ .

Solve the questions and the answers should be in English

Step-by-step solution

Apply the power rule of logarithms, $a \log_b c = \log_b c^a$ , to both sides of the equation.

Since the logarithms have the same base and are equal, their arguments must be equal.

Take the common logarithm (base 10) on both sides.

Apply the power rule of logarithms again.

Solve for $m$ .

Substitute the first equation ( $y = 2 \log_2 x$ ) into the second equation.

Use the product rule of logarithms, $\log_b (AB) = \log_b A + \log_b B$ , on the right side.

Rearrange the equation to isolate $\log_2 x$ .

Convert the logarithmic equation to an exponential equation using the definition $\log_b A = C \implies A = b^C$ .

Simplify the expression for $x$ .

From the equation $x+y=1$ , express $y$ in terms of $x$ .

Substitute this expression for $y$ into the equation $2^x = 3^y$ .

Take the common logarithm (base 10, denoted as $\log$ ) on both sides of the equation.

Apply the power rule of logarithms, $\log A^B = B \log A$ , to both sides.

Distribute $\log 3$ on the right side.

Gather all terms containing $x$ on one side of the equation.

Factor out $x$ from the terms on the left side.

Use the product rule of logarithms, $\log A + \log B = \log (AB)$ , on the expression in the parenthesis.

Solve for $x$ .

Still stuck on this one?

More English Questions

Solve the questions and the answers should be in English

Step-by-step solution

Apply the power rule of logarithms, alog⁡bc=log⁡bcaa \log_b c = \log_b c^aalogb​c=logb​ca, to both sides of the equation.

Since the logarithms have the same base and are equal, their arguments must be equal.

Take the common logarithm (base 10) on both sides.

Apply the power rule of logarithms again.

Solve for mmm.

Substitute the first equation (y=2log⁡2xy = 2 \log_2 xy=2log2​x) into the second equation.

Use the product rule of logarithms, log⁡b(AB)=log⁡bA+log⁡bB\log_b (AB) = \log_b A + \log_b Blogb​(AB)=logb​A+logb​B, on the right side.

Rearrange the equation to isolate log⁡2x\log_2 xlog2​x.

Convert the logarithmic equation to an exponential equation using the definition log⁡bA=C ⟹ A=bC\log_b A = C \implies A = b^Clogb​A=C⟹A=bC.

Simplify the expression for xxx.

From the equation x+y=1x+y=1x+y=1, express yyy in terms of xxx.

Substitute this expression for yyy into the equation 2x=3y2^x = 3^y2x=3y.

Take the common logarithm (base 10, denoted as log⁡\loglog) on both sides of the equation.

Apply the power rule of logarithms, log⁡AB=Blog⁡A\log A^B = B \log AlogAB=BlogA, to both sides.

Distribute log⁡3\log 3log3 on the right side.

Gather all terms containing xxx on one side of the equation.

Factor out xxx from the terms on the left side.

Use the product rule of logarithms, log⁡A+log⁡B=log⁡(AB)\log A + \log B = \log (AB)logA+logB=log(AB), on the expression in the parenthesis.

Solve for xxx.

Still stuck on this one?

More English Questions

Apply the power rule of logarithms, $a \log_b c = \log_b c^a$ , to both sides of the equation.

Solve for $m$ .

Substitute the first equation ( $y = 2 \log_2 x$ ) into the second equation.

Use the product rule of logarithms, $\log_b (AB) = \log_b A + \log_b B$ , on the right side.

Rearrange the equation to isolate $\log_2 x$ .

Convert the logarithmic equation to an exponential equation using the definition $\log_b A = C \implies A = b^C$ .

Simplify the expression for $x$ .

From the equation $x+y=1$ , express $y$ in terms of $x$ .

Substitute this expression for $y$ into the equation $2^x = 3^y$ .

Take the common logarithm (base 10, denoted as $\log$ ) on both sides of the equation.

Apply the power rule of logarithms, $\log A^B = B \log A$ , to both sides.

Distribute $\log 3$ on the right side.

Gather all terms containing $x$ on one side of the equation.

Factor out $x$ from the terms on the left side.

Use the product rule of logarithms, $\log A + \log B = \log (AB)$ , on the expression in the parenthesis.

Solve for $x$ .