// Numbas version: exam_results_page_options {"name": "Logs: resulting in negatives", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "variable_groups": [], "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": ""}, "tags": ["logarithms", "logs", "Logs", "negative indices", "negative powers"], "functions": {}, "name": "Logs: resulting in negatives", "ungrouped_variables": ["one", "two", "small", "tens", "num1", "num2"], "parts": [{"variableReplacementStrategy": "originalfirst", "steps": [{"variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "information", "prompt": "

When learning index laws you would have seen that
\\[x^{-n}=\\frac{1}{x^n}.\\]

Indices and logs are intimately related, ensure you revise your index laws.

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Recall that we can convert negative indices to fractions

$\\var{num1}^{-1}$ = [[0]]

$\\var{num2}^{-2}$ = [[1]]

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To determine the value of $\\log_b(a)$, you can think \"$b$ to the what equals $a$\", and that will be your answer.

To determine $\\log_{\\var{one[0]}}\\left(\\frac{1}{\\var{one[0]}}\\right)$, realise $\\var{one[0]}^{-1}=\\frac{1}{\\var{one[0]}}$ and so $\\log_{\\var{one[0]}}\\left(\\frac{1}{\\var{one[0]}}\\right)=-1$.

Using the definition and your times tables determine the following:

$\\log_{\\var{one[0]}}\\left(\\frac{1}{\\var{one[0]}}\\right)$ = [[0]]

To determine the value of $\\log_b(a)$, you can think \"$b$ to the what equals $a$\", and that will be your answer.

To determine $\\log_{\\var{two[0]}}\\left(\\frac{1}{\\var{two[0]^2}}\\right)$, realise $\\var{two[0]}^{-2}=\\frac{1}{\\var{two[0]^2}}$ and so $\\log_{\\var{two[0]}}\\left(\\frac{1}{\\var{two[0]^2}}\\right)=-2$.

Using the definition and your times tables determine the following:

$\\log_{\\var{two[0]}}\\left(\\frac{1}{\\var{two[0]^2}}\\right)$ = [[0]]

To determine the value of $\\log_b(a)$, you can think \"$b$ to the what equals $a$\", and that will be your answer.

To determine $\\log_{\\var{small[0]}}\\left(\\frac{1}{\\var{small[0]^small[1]}}\\right)$, realise $\\var{small[0]}^{-\\var{small[1]}}=\\frac{1}{\\var{small[0]^small[1]}}$ and so $\\log_{\\var{small[0]}}\\left(\\frac{1}{\\var{small[0]^small[1]}}\\right)=-\\var{small[1]}$.

Using the definition and your times tables determine the following:

$\\log_{\\var{small[0]}}\\left(\\frac{1}{\\var{small[0]^small[1]}}\\right)$ = [[0]]

To determine the value of $\\log_b(a)$, you can think \"$b$ to the what equals $a$\", and that will be your answer.

To determine $\\log_{\\var{tens[0]}}(\\var{1/tens[0]^tens[1]})$, realise $\\var{tens[0]}^{-\\var{tens[1]}}=\\var{1/tens[0]^tens[1]}$ and so $\\log_{\\var{tens[0]}}(\\var{1/tens[0]^tens[1]})=-\\var{tens[1]}$.

Recall that $10^{-n}$ is the same as a decimal with zeros everywhere except a $1$ at the $n$th decimal place.

Using the definition and your times tables determine the following:

$\\log_{\\var{tens[0]}}(\\var{1/tens[0]^tens[1]})$ = [[0]]

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The following should be completed without the use of a calculator.

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