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

Level 1

This math topic focuses on advanced logarithm functions, particularly using the power property of logarithms to isolate exponents and solve equations. The problems involve equating two binomials, each raised to a logarithmic power, in order to solve for a variable. Most questions require simplifying expressions using logarithmic identities and algebraically manipulating them to find the variable. This set of problems helps in understanding the application of logarithmic functions in various algebraic contexts.

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

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

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

A LaTex expression showing 7 to the power of (n + 8) = 9 to the power of (n + 4)

a $A LaTex expression showing n=\frac{ 4\ln{7} - 8\ln{9}}{ \ln{9} - \ln{7}}$

b $A LaTex expression showing n=\frac{ 4\ln{9} - 8\ln{7}}{ \ln{7} - 1\ln{9}}$

c $A LaTex expression showing n=\frac{ \ln{9} - \ln{7}}{ 8\ln{7} - 4\ln{9}}$

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

A LaTex expression showing 4 to the power of (z - 3) = 6 to the power of (z - 3)

a $A LaTex expression showing z=\frac{ \ln{6} - \ln{4}}{ -3\ln{4} + 3\ln{6}}$

b $A LaTex expression showing z=\frac{ -3\ln{4} + 3\ln{6}}{ \ln{6} - \ln{4}}$

c $A LaTex expression showing z=\frac{ -3\ln{6} + 3\ln{4}}{ \ln{4} - 1\ln{6}}$

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

A LaTex expression showing 8 to the power of (r - 2) = 10 to the power of (r + 2)

a $A LaTex expression showing r=\frac{ 2\ln{10} + 2\ln{8}}{ \ln{8} - 1\ln{10}}$

b $A LaTex expression showing r=\frac{ \ln{10} - \ln{8}}{ -2\ln{8} - 2\ln{10}}$

c $A LaTex expression showing r=\frac{ 2\ln{8} + 2\ln{10}}{ \ln{10} - \ln{8}}$

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

A LaTex expression showing 9 to the power of (y - 5) = 2 to the power of (y - 9)

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

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

c $A LaTex expression showing y=\frac{ -9\ln{2} + 5\ln{9}}{ \ln{9} - 1\ln{2}}$

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

A LaTex expression showing 6 to the power of (x + 3) = 5 to the power of (x + 1)

a $A LaTex expression showing x=\frac{ 1\ln{6} - 3\ln{5}}{ \ln{5} - \ln{6}}$

b $A LaTex expression showing x=\frac{ \ln{5} - \ln{6}}{ 3\ln{6} - \ln{5}}$

c $A LaTex expression showing x=\frac{ \ln{5} - 3\ln{6}}{ \ln{6} - 1\ln{5}}$

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

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

a $A LaTex expression showing r=\frac{ 4\ln{9} - 7\ln{4}}{ \ln{4} - 1\ln{9}}$

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

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

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

A LaTex expression showing 9 to the power of (y - 7) = 2 to the power of (y - 2)

a $A LaTex expression showing y=\frac{ \ln{2} - \ln{9}}{ -7\ln{9} + 2\ln{2}}$

b $A LaTex expression showing y=\frac{ -2\ln{2} + 7\ln{9}}{ \ln{9} - 1\ln{2}}$

c $A LaTex expression showing y=\frac{ -2\ln{9} + 7\ln{2}}{ \ln{2} - \ln{9}}$

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

A LaTex expression showing 8 to the power of (m - 8) = 6 to the power of (m + 9)

a $A LaTex expression showing m=\frac{ 9\ln{8} + 8\ln{6}}{ \ln{6} - \ln{8}}$

b $A LaTex expression showing m=\frac{ \ln{6} - \ln{8}}{ -8\ln{8} - 9\ln{6}}$

c $A LaTex expression showing m=\frac{ 9\ln{6} + 8\ln{8}}{ \ln{8} - 1\ln{6}}$