Logarithm Algebra (Power Property) - Isolote Exponent, Two Binomials (Coefficient 1) to Partial Answer

Level 1

This topic focuses on utilizing the power property of logarithms to simplify equations. It covers scenarios where exponential functions with similar exponents but different bases are set equal to each other. Each problem provides multiple choice solutions where the key task is to apply logarithms effectively to isolate variables within exponent terms. This is an advanced exercise in understanding and applying logarithm functions, particularly useful for deciphering how to handle equations involving exponential terms with coefficients and manipulated variables.

Work on practice problems directly here, or download the printable pdf worksheet to practice offline.

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Logarithm Algebra (Power Property) - Isolote Exponent, Two Binomials (Coefficient 1) to Partial Answer Worksheet

Logarithm Algebra (Power Property) - Isolote Exponent, Two Binomials ...

Use the power rule to simplify this equation

A LaTex expression showing 2 to the power of (q + 4) = 3 to the power of (q + 4)

a $A LaTex expression showing (4q + 1)\ln{2} = (4q + 1)\ln{3}$

b $A LaTex expression showing (q + 4)\ln{2} = (q + 4)\ln{3}$

c $A LaTex expression showing (q - 4)\ln{2} = (q - 4)\ln{3}$

Use the power rule to simplify this equation

A LaTex expression showing 4 to the power of (p + 7) = 6 to the power of (p + 7)

a $A LaTex expression showing (p + 7)\ln{4} = (p + 7)\ln{6}$

b $A LaTex expression showing (p - 7)\ln{4} = (p - 7)\ln{6}$

c $A LaTex expression showing (7p + 1)\ln{4} = (7p + 1)\ln{6}$

Use the power rule to simplify this equation

A LaTex expression showing 3 to the power of (z - 5) = 6 to the power of (z - 2)

a $A LaTex expression showing (z - 5)\ln{3} = (z - 2)\ln{6}$

b $A LaTex expression showing (z + 5)\ln{3} = (z + 2)\ln{6}$

c $A LaTex expression showing (-5z + 1)\ln{3} = (-2z + 1)\ln{6}$

Use the power rule to simplify this equation

A LaTex expression showing 9 to the power of (y + 1) = 7 to the power of (y + 4)

a $A LaTex expression showing (y - 1)\ln{9} = (y - 4)\ln{7}$

b $A LaTex expression showing (y + 1)\ln{9} = (4y + 1)\ln{7}$

c $A LaTex expression showing (y + 1)\ln{9} = (y + 4)\ln{7}$

Use the power rule to simplify this equation

A LaTex expression showing 2 to the power of (r + 5) = 10 to the power of (r + 9)

a $A LaTex expression showing (r + 5)\ln{2} = (r + 9)\ln{10}$

b $A LaTex expression showing (5r + 1)\ln{2} = (9r + 1)\ln{10}$

c $A LaTex expression showing (r - 5)\ln{2} = (r - 9)\ln{10}$

Use the power rule to simplify this equation

A LaTex expression showing 9 to the power of (q + 5) = 7 to the power of (q - 4)

a $A LaTex expression showing (5q + 1)\ln{9} = (-4q + 1)\ln{7}$

b $A LaTex expression showing (q + 5)\ln{9} = (q - 4)\ln{7}$

c $A LaTex expression showing (q - 5)\ln{9} = (q + 4)\ln{7}$

Use the power rule to simplify this equation

A LaTex expression showing 9 to the power of (w - 8) = 7 to the power of (w - 6)

a $A LaTex expression showing (w + 8)\ln{9} = (w + 6)\ln{7}$

b $A LaTex expression showing (-8w + 1)\ln{9} = (-6w + 1)\ln{7}$

c $A LaTex expression showing (w - 8)\ln{9} = (w - 6)\ln{7}$

Use the power rule to simplify this equation

A LaTex expression showing 2 to the power of (y - 6) = 10 to the power of (y - 5)

a $A LaTex expression showing (y + 6)\ln{2} = (y + 5)\ln{10}$

b $A LaTex expression showing (y - 6)\ln{2} = (y - 5)\ln{10}$

c $A LaTex expression showing (-6y + 1)\ln{2} = (-5y + 1)\ln{10}$