HiSET Exponential and Logarithmic Equations — practice questions

1. Condense the left side with the product rule into one logarithm. 2. That gives log base 2 of the product x times the bracket (x minus 2), equal to 3. 3. Rewrite in exponential form, so x times the bracket (x minus 2) equals 2 cubed, which is 8. 4. Expanding gives x squared minus 2x equal to 8. 5. Writing that in standard form gives x squared minus 2x minus 8 equal to 0. 6. Factoring gives the brackets (x minus 4) and (x plus 2), so the candidates are 4 and negative 2. 7. Check each one, since a logarithm accepts only positive arguments. 8. At x equal to 4 the arguments are 4 and 2, both positive, and log base 2 of 4 plus log base 2 of 2 is 2 plus 1, which is 3. 9. At x equal to negative 2 the first argument is negative, so that candidate has to be thrown out. 10. So the value 4 only survives the check, and keeping both candidates or rejecting both would be wrong. _Source: OpenStax Intermediate Algebra (CC BY 4.0), section 10.5 Solve Exponential and Logarithmic Equations_

1. The two sides cannot be written as powers of a common base, so a logarithm is needed. 2. The exponential is already alone, so take the logarithm of both sides. 3. That gives log of 7 to the power x equal to log of 30. 4. The power rule brings the exponent down in front, giving x times log 7 equal to log 30. 5. Divide both sides by log 7 to isolate x. 6. So x equals log 30 divided by log 7. 7. Flipping the fraction would answer the different question of what power of 30 gives 7. 8. Subtracting the logarithms would answer a division inside a single logarithm instead. 9. Dividing 30 by 7 ignores the exponent entirely and gives a value under 5, which is far too large. _Source: OpenStax Intermediate Algebra (CC BY 4.0), section 10.5 Solve Exponential and Logarithmic Equations_

_Source: OpenStax Intermediate Algebra (CC BY 4.0), section 10.3 Evaluate and Graph Logarithmic Functions_