GMAT Math : Graphing a logarithm

Study concepts, example questions & explanations for GMAT Math

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Example Questions

Example Question #818 : Geometry

Let \(\displaystyle (x,y)\) be the point of intersection of the graphs of these two equations:

\(\displaystyle 2 \log x+3 \log y = 12\)

\(\displaystyle 5 \log x- 2 \log y = 11\)

Evaluate \(\displaystyle x\).

Possible Answers:

\(\displaystyle x =\frac{1}{ 1,000}\)

\(\displaystyle x =\frac{1}{ 1 00}\)

\(\displaystyle x = 1,000\)

\(\displaystyle x = 100\)

The system has no solution.

Correct answer:

\(\displaystyle x = 1,000\)

Explanation:

Substitute \(\displaystyle u\) and \(\displaystyle v\) for \(\displaystyle \log x\) and \(\displaystyle \log y\), respectively, and solve the resulting system of linear equations:

\(\displaystyle 2 u+3 v = 12\)

\(\displaystyle 5 u- 2 v = 11\)

Multiply the first equation by 2, and the second by 3, on both sides, then add:

\(\displaystyle 4 u+6 v = 24\)

\(\displaystyle \underline{15 u- 6 v = 33}\)

\(\displaystyle 19u\)            \(\displaystyle =57\)

\(\displaystyle u = 57 \div 19 = 3\)

Back-solve:

\(\displaystyle 2 (3)+3 v = 12\)

\(\displaystyle 6+3 v = 12\)

\(\displaystyle 3v = 6\)

\(\displaystyle v = 2\)

We need to find both \(\displaystyle x\) and \(\displaystyle y\) to ensure a solution exists. By substituting back:

\(\displaystyle u = 3\)

\(\displaystyle \log x = 3\)

\(\displaystyle x=10^{3} = 1,000\)

and

\(\displaystyle v=2\)

\(\displaystyle \log y = 2\)

\(\displaystyle y=10^{2} = 100\)

We check this solution in both equations:

\(\displaystyle 2 \log x+3 \log y = 12\)

\(\displaystyle 2 \log 1,000+3 \log 100 = 12\)

\(\displaystyle 2 (3)+3 (2)= 12\)

\(\displaystyle 6+6 = 12\) - true.

 

\(\displaystyle 5 \log x- 2 \log y = 11\)

\(\displaystyle 5 \log 1,000- 2 \log 100 = 11\)

\(\displaystyle 5 (3) -2 (2) = 11\)

\(\displaystyle 15-4 = 11\) - true.

 

\(\displaystyle (1000, 100)\) is the solution, and \(\displaystyle x=1,000\), the correct choice.

 

 

Example Question #211 : Coordinate Geometry

The graph of function \(\displaystyle f\) has vertical asymptote \(\displaystyle x = 5\). Which of the following could give a definition of \(\displaystyle f\) ?

Possible Answers:

\(\displaystyle f(x) =7\log (x-3)+5\)

\(\displaystyle f(x) =3\log (x-7)-5\)

\(\displaystyle f(x) = -5 \log (x-7)-3\)

\(\displaystyle f(x) = 5 \log (x-3)-7\)

\(\displaystyle f(x) = 3 \log (x-5)+7\)

Correct answer:

\(\displaystyle f(x) = 3 \log (x-5)+7\)

Explanation:

Given the function \(\displaystyle f(x) = A \log (x-B)+C\), the vertical asymptote can be found by observing that a logarithm cannot be taken of a number that is not positive. Therefore, it must hold that \(\displaystyle x-B >0\), or, equivalently, \(\displaystyle x>B\) and that the graph of \(\displaystyle f\) will never cross the vertical line \(\displaystyle x=B\). That makes \(\displaystyle x=B\) the vertical asymptote, so it follows that the graph with vertical asymptote \(\displaystyle x = 5\) will have \(\displaystyle 5\) in the \(\displaystyle B\) position. The only choice that meets this criterion is

\(\displaystyle f(x) = 3 \log (x-5)+7\)

Example Question #221 : Coordinate Geometry

The graph of a function \(\displaystyle f\) has \(\displaystyle x\)-intercept \(\displaystyle (64,0)\). Which of the following could be the definition of \(\displaystyle f\) ?

Possible Answers:

\(\displaystyle f(x)= \log_{2}x-6\)

\(\displaystyle f(x)= \log_{4}x-3\)

All of the other choices are correct.

\(\displaystyle f(x)= \log_{16}x- \frac{3}{2}\)

\(\displaystyle f(x)= \log_{8}x-2\)

Correct answer:

All of the other choices are correct.

Explanation:

All of the functions are of the form \(\displaystyle f(x) = \log_{b}x - C\). To find the \(\displaystyle x\)-intercept of such a function, we can set \(\displaystyle f(x) = 0\) and solve for \(\displaystyle x\):

\(\displaystyle \log_{b}x - C = 0\)

\(\displaystyle \log_{b}x =C\)

\(\displaystyle b^{C}= x\)

Since we are looking for a function whose graph has \(\displaystyle x\)-intercept \(\displaystyle (64,0)\), the equation here becomes \(\displaystyle b^{C}= 64\), and we can examine each of the functions by finding the value of \(\displaystyle b^{C}\).

\(\displaystyle f(x)= \log_{8}x-2\):

 \(\displaystyle b= 8, c= 2\)

\(\displaystyle b^{C}= 8^{2} = 64\)

 

\(\displaystyle f(x)= \log_{2}x-6\)

\(\displaystyle b = 2,c=6\)

\(\displaystyle b^{C}= 2^{6} = 64\)

 

\(\displaystyle f(x)= \log_{4}x-3\)

\(\displaystyle b=4,c=3\)

\(\displaystyle b^{C}=4^{3} = 64\)

 

\(\displaystyle f(x)= \log_{16}x- \frac{3}{2}\)

\(\displaystyle b=16,c=\frac{3}{2}\)

\(\displaystyle b^{C}= 16^{ \frac{3}{2}} = \left ( \sqrt{16} \right )^{3}= 4^{3}= 64\)

 

All four choices fit the criterion.

Example Question #1061 : Problem Solving Questions

The graph of a function \(\displaystyle f\) has \(\displaystyle y\,\)-intercept \(\displaystyle (0,8)\). Which of the following could be the definition of \(\displaystyle f\) ?

Possible Answers:

\(\displaystyle f(x)= \log _{2}x+6\)

\(\displaystyle f(x)= \log _{2}(x+8) +6\)

\(\displaystyle f(x)= \log _{2}(x+2) +6\)

\(\displaystyle f(x)= \log _{2}(x+4) +6\)

\(\displaystyle f(x)= \log _{2}(x+1) +6\)

Correct answer:

\(\displaystyle f(x)= \log _{2}(x+4) +6\)

Explanation:

All of the functions take the form 

\(\displaystyle f(x)= \log _{2}(x+A) +6\)

for some integer \(\displaystyle A\). To find the choice that has \(\displaystyle y\,\)-intercept \(\displaystyle (0,8)\), set \(\displaystyle x = 0\) and \(\displaystyle f(x)= 8\), and solve for \(\displaystyle A\):

\(\displaystyle 8= \log _{2}(0+A) +6\)

\(\displaystyle 8= \log _{2}A +6\)

\(\displaystyle 2= \log _{2}A\)

In exponential form:

\(\displaystyle A =2^{2}=4\)

The correct choice is \(\displaystyle f(x)= \log _{2}(x+4) +6\).

Example Question #824 : Geometry

Define a function \(\displaystyle g\) as follows:

\(\displaystyle g(x) =-3 \log_{8}(x+2)+ 2\)

Give the \(\displaystyle x\)-intercept of the graph of \(\displaystyle g\).

Possible Answers:

\(\displaystyle (-2, 0)\)

\(\displaystyle (6,0)\)

\(\displaystyle \left ( -2 + 16 \sqrt{2}, 0 \right )\)

\(\displaystyle (2,0)\)

\(\displaystyle \left ( -2 - 16 \sqrt{2}, 0 \right )\)

Correct answer:

\(\displaystyle (2,0)\)

Explanation:

Set \(\displaystyle g(x) =0\) and evaluate \(\displaystyle x\) to find the \(\displaystyle x\)-coordinate of the \(\displaystyle x\)-intercept.

\(\displaystyle g(x) =-3 \log_{8}(x+2)+ 2 = 0\)

\(\displaystyle -3 \log_{8}(x+2)= - 2\)

\(\displaystyle \log_{8}(x+2)= \frac{-2}{-3}= \frac{2}{3}\)

Rewrite in exponential form:

\(\displaystyle x+ 2 = 8^{\frac{2}{3}}\)

\(\displaystyle x+ 2 =\left ( \sqrt[3]{8} \right )^{2} = 2^{2} = 4\)

\(\displaystyle x=2\).

The \(\displaystyle x\)-intercept is \(\displaystyle (2,0)\).

Example Question #825 : Geometry

Define functions \(\displaystyle f\) and \(\displaystyle g\) as follows:

\(\displaystyle f(x) = 2 \log (x-4)\)

\(\displaystyle g(x) = \log (x+3)+ \log (x-2)\)

Give the \(\displaystyle y\)-coordinate of a point at which the graphs of the functions intersect.

Possible Answers:

\(\displaystyle 2 \frac{4}{9}\)

\(\displaystyle \log 2\frac{34}{81}\)

The graphs of \(\displaystyle f\) and \(\displaystyle g\) do not intersect.

\(\displaystyle \log 1\frac{5}{9}\)

\(\displaystyle 5\)

Correct answer:

The graphs of \(\displaystyle f\) and \(\displaystyle g\) do not intersect.

Explanation:

Since \(\displaystyle a \log N = \log( N^{a})\), the definition of \(\displaystyle f\) can be rewritten as follows:

\(\displaystyle f(x) = 2 \log (x-4)\)

\(\displaystyle f(x) = \log (x-4) ^{2}\)

Since \(\displaystyle \log M + \log N = \log MN\), the definition of \(\displaystyle g\) can be rewritten as follows:

\(\displaystyle g(x) = \log (x+3)+ \log (x-2)\)

\(\displaystyle g(x) = \log (x+3) (x-2)\)

First, we need to find the \(\displaystyle x\)-coordinate of the point at which the graphs of \(\displaystyle f\) and \(\displaystyle g\) meet by setting 

\(\displaystyle f(x)= g(x)\)

\(\displaystyle \log (x-4) ^{2} = \log (x+3) (x-2)\)

Since the common logarithms of the two polynomials are equal, we can set the polynomials themselves equal, then solve:

\(\displaystyle (x-4) ^{2} = (x+3) (x-2)\)

\(\displaystyle x ^{2}- 8x+16 = x^{2} +x-6\)

\(\displaystyle x ^{2}- 8x+16 - x^{2} -x - 16 = x^{2} +x-6 - x^{2} -x - 16\)

\(\displaystyle -9x = - 22\)

\(\displaystyle x = \frac{22}{9} = 2\frac{4}{9}\)

However, if we evaluate \(\displaystyle f \left ( 2\frac{4}{9} \right )\), the expression becomes

\(\displaystyle f(x) = 2 \log (x-4)\)

\(\displaystyle f\left ( 2\frac{4}{9} \right ) = 2\log \left ( 2\frac{4}{9} -4\right ) = 2 \log \left ( -1\frac{5}{9} \right )\),

which is undefined, since a negative number cannot have a logarithm.

Consequently, the two graphs do not intersect.

 

Example Question #826 : Geometry

The graph of a function \(\displaystyle f\) has \(\displaystyle x\)-intercept \(\displaystyle \left ( \frac{1}{32}, 0 \right )\). Which of the following could be the definition of \(\displaystyle f\) ?

Possible Answers:

\(\displaystyle f(x)= \log_{5}x-2\)

None of the other responses gives a correct answer.

\(\displaystyle f(x)= \log_{2}x+5\)

\(\displaystyle f(x)= \log_{2}x-5\)

\(\displaystyle f(x)= \log_{5}x+2\)

Correct answer:

\(\displaystyle f(x)= \log_{2}x+5\)

Explanation:

All of the functions are of the form \(\displaystyle f(x) = \log_{b}x - C\). To find the \(\displaystyle x\)-intercept of a function \(\displaystyle f(x) = \log_{b}x - C\),  we can set \(\displaystyle f(x) = 0\) and solve for \(\displaystyle x\):

\(\displaystyle \log_{b}x - C = 0\)

\(\displaystyle \log_{b}x =C\)

\(\displaystyle b^{C}= x\).

Since we are looking for a function whose graph has \(\displaystyle x\)-intercept \(\displaystyle \left ( \frac{1}{32}, 0 \right )\), the equation here becomes \(\displaystyle b^{C}= \frac{1}{32}\), and we can examine each of the functions by finding the value of \(\displaystyle b^{C}\) and seeing which case yields this result.

 

\(\displaystyle f(x)= \log_{5}x-2\):

\(\displaystyle b = 5, C=2\)

\(\displaystyle b^{C} = 5^{2} = 25\)

 

\(\displaystyle f(x)= \log_{2}x-5\):

\(\displaystyle b = 2, C=5\)

\(\displaystyle b^{C} = 2^{5} = 32\)

 

\(\displaystyle f(x)= \log_{5}x+2\):

\(\displaystyle b = 5, C=-2\)

\(\displaystyle b^{C} = 5^{-2} = \frac{1}{5^{2}}= \frac{1}{25}\)

 

\(\displaystyle f(x)= \log_{2}x+5\):

\(\displaystyle b = 2, C=-5\)

\(\displaystyle b^{C} = 2^{-5} = \frac{1}{2^{5}}= \frac{1}{32}\)

The graph of \(\displaystyle f(x)= \log_{2}x+5\) has \(\displaystyle x\)-intercept \(\displaystyle \left ( \frac{1}{32}, 0 \right )\) and is the correct choice.

Example Question #827 : Geometry

Define a function \(\displaystyle g\) as follows:

\(\displaystyle g(x) =2 \log_{9}(x+3)- 3\)

A line passes through the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts of the graph of \(\displaystyle g\). Give the equation of the line.

Possible Answers:

\(\displaystyle y = 12x+24\)

\(\displaystyle y= -\frac{1}{12} x +2\)

\(\displaystyle y= \frac{1}{12} x -2\)

\(\displaystyle y =- 12x+30\)

\(\displaystyle y= -\frac{1}{12} x -4\)

Correct answer:

\(\displaystyle y= \frac{1}{12} x -2\)

Explanation:

The \(\displaystyle x\)-intercept of the graph of \(\displaystyle g\) can befound by setting \(\displaystyle g(x) = 0\) and solving for \(\displaystyle x\):

\(\displaystyle 2 \log_{9}(x+3)- 3 = 0\)

\(\displaystyle 2 \log_{9}(x+3) = 3\)

\(\displaystyle \log_{9}(x+3) = \frac{3}{2}\)

Rewritten in exponential form:

\(\displaystyle x+3 = 9 ^{\frac{3}{2}}\)

\(\displaystyle x+3 = (\sqrt{9} )^{3}\)

\(\displaystyle x+3 = 3^{3}\)

\(\displaystyle x+3 = 27\)

\(\displaystyle x = 24\)

The \(\displaystyle x\)-intercept of the graph of \(\displaystyle g\) is \(\displaystyle (24,0)\).

 

The \(\displaystyle y \,\)-intercept of the graph of \(\displaystyle g\) can be found by evaluating \(\displaystyle g(0):\)

\(\displaystyle g(x) =2 \log_{9}(x+3)- 3\)

\(\displaystyle g(0) =2 \log_{9}(0+3)- 3\)

\(\displaystyle =2 \log_{9}3- 3\)

\(\displaystyle = \log_{9}3^{2}- 3\)

\(\displaystyle = \log_{9}9- 3\)

\(\displaystyle = 1-3\)

\(\displaystyle = -2\)

The \(\displaystyle y \,\)-intercept of the graph of \(\displaystyle g\) is \(\displaystyle (0, -2)\).

 

If \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) are the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of a line, the slope of the line is \(\displaystyle m=-\frac{b}{a}\). Substituting \(\displaystyle a = 24\) and \(\displaystyle b = -2\), this is

\(\displaystyle m=-\frac{-2}{24} = \frac{1}{12}\).

Setting \(\displaystyle m= \frac{1}{12}\) and \(\displaystyle b = -2\) in the slope-intercept form of the equation of a line:

\(\displaystyle y = mx+b\)

\(\displaystyle y= \frac{1}{12} x -2\)

Example Question #1061 : Problem Solving Questions

Define functions \(\displaystyle f\) and \(\displaystyle g\) as follows:

\(\displaystyle f(x) = 2 \log (x+5)\)

\(\displaystyle g(x) = \log (2x+13)\)

Give the \(\displaystyle y\)-coordinate of a point at which the graphs of the functions intersect.

Possible Answers:

\(\displaystyle \log 6\)

The graphs of \(\displaystyle f\) and \(\displaystyle g\) do not intersect.

\(\displaystyle 0\)

\(\displaystyle -2\)

\(\displaystyle \log 9\)

Correct answer:

\(\displaystyle \log 9\)

Explanation:

Since \(\displaystyle a \log N= \log( N^{a})\), the definition of \(\displaystyle f\) can be rewritten as follows:

 \(\displaystyle f(x) = \log \left [(x+5)^{2} \right ]\)

Find the \(\displaystyle x\)-coordinate of the point at which the graphs of \(\displaystyle f\) and \(\displaystyle g\) meet by setting 

\(\displaystyle f(x)= g(x)\)

\(\displaystyle \log \left [(x+5)^{2} \right ]= \log (2x+13)\)

Since the common logarithms of the two polynomials are equal, we can set the polynomials themselves equal, then solve:

\(\displaystyle (x+5)^{2} = 2x+13\)

\(\displaystyle x^{2} +10x+25= 2x+13\)

\(\displaystyle x^{2} +10x+25- 2x - 13 = 2x+13 - 2x - 13\)

\(\displaystyle x^{2} +8x+12= 0\)

The quadradic trinomial can be "reverse-FOILed" by noting that 2 and 6 have  product 12 and sum 8:

\(\displaystyle (x+2)(x+6 ) = 0\)

Either \(\displaystyle x+2 = 0\), in which case \(\displaystyle x=-2\)

or

\(\displaystyle x+6 = 0\), in which case \(\displaystyle x = -6\)

Note, however, that we can eliminate \(\displaystyle x = -6\) as a possible \(\displaystyle x\)-value, since

\(\displaystyle f(-6) = 2 \log (-6+5) = 2 \log (-1)\),

an undefined quantity since negative numbers do not have logarithms. 

Since 

\(\displaystyle f(-2) = 2 \log (-2+5) = 2 \log 3 = \log (3^{2}) = \log 9\)

and 

\(\displaystyle g(-2) = \log [2(-2)+13] = \log (-4+13) = \log 9\),

\(\displaystyle -2\) is the correct \(\displaystyle x\)-value, and \(\displaystyle \log 9\) is the correct \(\displaystyle y\)-value.

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