Logarithm Algebra (Power Property) - Isolote Exponent, One Binomial (Coefficient N) to Answer

Level 1

This math topic focuses on practicing the power property of logarithms. Students apply this property to isolate variables within exponential equations involving binomials as coefficients. The problems involve equating expressions of different bases, then taking logarithms and solving for the variables (e.g., 'q', 'z', 'x', 'm', 't', 'n', 'r'). Each question includes multiple-choice answers where students must manipulate logarithmic expressions to identify the correct solution. This topic is part of an advanced unit on logarithm functions.

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Logarithm Algebra (Power Property) - Isolote Exponent, One Binomial (Coefficient N) to Answer Worksheet

Logarithm Algebra (Power Property) - Isolote Exponent, One Binomial (...

Use the power rule to simplify this and solve for 'q'

A LaTex expression showing 6 to the power of (-6q + 3) = 3 to the power of (3q )

a $A LaTex expression showing q=\frac{ -3\ln{6}}{ -6\ln{6} - 3\ln{3}}$

b $A LaTex expression showing q=\frac{ -3\ln{3}}{ -6\ln{3} - 3\ln{6}}$

c $A LaTex expression showing q=\frac{ 3\ln{3} + 6\ln{6}}{ 3\ln{6} - 2\ln{3}}$

Use the power rule to simplify this and solve for 'z'

A LaTex expression showing 7 to the power of (-7z - 7) = 6 to the power of (-7z )

a $A LaTex expression showing z=\frac{ -7\ln{6} + 7\ln{7}}{ -7\ln{7} - 2\ln{6}}$

b $A LaTex expression showing z=\frac{ 7\ln{6}}{ -7\ln{6} + 7\ln{7}}$

c $A LaTex expression showing z=\frac{ 7\ln{7}}{ -7\ln{7} + 7\ln{6}}$

Use the power rule to simplify this and solve for 'x'

A LaTex expression showing 5 to the power of (-1x - 2) = 10 to the power of (-2x )

a $A LaTex expression showing x=\frac{ 2\ln{5}}{ -1\ln{5} + 2\ln{10}}$

b $A LaTex expression showing x=\frac{ 2\ln{10}}{ -1\ln{10} + 2\ln{5}}$

c $A LaTex expression showing x=\frac{ -2\ln{10} + \ln{5}}{ -2\ln{5} - 2\ln{10}}$

Use the power rule to simplify this and solve for 'm'

A LaTex expression showing 3 to the power of (-8m + 9) = 2 to the power of (8m )

a $A LaTex expression showing m=\frac{ -9\ln{3}}{ -8\ln{3} - 8\ln{2}}$

b $A LaTex expression showing m=\frac{ 8\ln{2} + 8\ln{3}}{ 9\ln{3} - 2\ln{2}}$

c $A LaTex expression showing m=\frac{ -9\ln{2}}{ -8\ln{2} - 8\ln{3}}$

Use the power rule to simplify this and solve for 't'

A LaTex expression showing 9 to the power of (-5t - 9) = 5 to the power of (9t )

a $A LaTex expression showing t=\frac{ 9\ln{5}}{ -5\ln{5} - 9\ln{9}}$

b $A LaTex expression showing t=\frac{ 9\ln{9}}{ -5\ln{9} - 9\ln{5}}$

c $A LaTex expression showing t=\frac{ 9\ln{5} + 5\ln{9}}{ -9\ln{9} - 2\ln{5}}$

Use the power rule to simplify this and solve for 'n'

A LaTex expression showing 2 to the power of (7n - 6) = 4 to the power of (-9n )

a $A LaTex expression showing n=\frac{ -9\ln{4} - 7\ln{2}}{ -6\ln{2} - 2\ln{4}}$

b $A LaTex expression showing n=\frac{ 6\ln{4}}{ 7\ln{4} + 9\ln{2}}$

c $A LaTex expression showing n=\frac{ 6\ln{2}}{ 7\ln{2} + 9\ln{4}}$

Use the power rule to simplify this and solve for 'r'

A LaTex expression showing 4 to the power of (9r + 9) = 10 to the power of (7r )

a $A LaTex expression showing r=\frac{ 7\ln{10} - 9\ln{4}}{ 9\ln{4} - 2\ln{10}}$

b $A LaTex expression showing r=\frac{ -9\ln{10}}{ 9\ln{10} - 7\ln{4}}$

c $A LaTex expression showing r=\frac{ -9\ln{4}}{ 9\ln{4} - 7\ln{10}}$

Use the power rule to simplify this and solve for 't'

A LaTex expression showing 2 to the power of (-5t - 2) = 8 to the power of (-8t )

a $A LaTex expression showing t=\frac{ -8\ln{8} + 5\ln{2}}{ -2\ln{2} - 2\ln{8}}$

b $A LaTex expression showing t=\frac{ 2\ln{2}}{ -5\ln{2} + 8\ln{8}}$

c $A LaTex expression showing t=\frac{ 2\ln{8}}{ -5\ln{8} + 8\ln{2}}$