// Numbas version: exam_results_page_options {"name": "Rosanna's copy of Logarithms: inverses", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"preamble": {"js": "", "css": ""}, "parts": [{"gaps": [{"showFeedbackIcon": true, "expectedvariablenames": [], "variableReplacements": [], "marks": 1, "answer": "x+{d}", "checkingtype": "absdiff", "notallowed": {"message": "

This expression can be simplified to not contain logs or exponentials.

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$\\var{b}^{\\log_{\\var{b}}(\\simplify{x+{d}})}=$ [[0]], for $x>\\var{-d}$.

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This expression can be simplified to not contain logs or exponentials.

$e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}=$ [[0]], for $\\simplify{x^{a}+{d}}>0$.

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This expression can be simplified to not contain logs or exponentials.

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$\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=$ [[0]]

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Simplify the following expressions:

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You might recall that $b^{\\log_{b}(a)}=a$ (for $a>0$), and therefore

\\[\\var{b}^{\\log_{\\var{b}}(\\simplify{x+{d}})}=\\simplify{x+{d}}, \\quad\\text{ for }x>\\var{-d}.\\]

We can think of the logarithm and the exponential (of the same base) cancelling each other out, or undoing each other since they are the inverse of each other. Some people might prefer to see more evidence of this, so here is a longer explanation:

Let $\\simplify{log(x+{d},{b})}=n$. Recall the definition of log says that this is equivalent to \\[\\simplify{{b}^n=x+{d}}.\\] Now substitute $\\simplify{log(x+{d},{b})}$ for $n$ into this equation (since we defined them to be equal) and we have

\\[\\simplify{{b}^(log(x+{d},{b}))=x+{d}}\\]

b) Recall that $\\ln$ denotes $\\log_e$.

Here is one way to do this simplification (use the log laws $b\\log(a)=\\log(a^b)$ and $b^{\\log_b(a)}=a$):

$\\begin{align}e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}&=e^{\\ln\\left(\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\right)}\\\\&=\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\end{align}$

Another way (use the index law $a^{bc}=(a^b)^c$ and $b^{\\log_b(a)}=a$):

$\\begin{align}e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}&=\\left(e^{\\ln(\\simplify{x^{a}+{d}})}\\right)^\\var[fractionNumbers]{c}\\\\&=\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\end{align}$

c) Recall that $\\ln$ denotes $\\log_e$.

To simplify $\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)$ we might think of the logarithm and the exponential (of the same base) cancelling each other out, or undoing each other since they are the inverse of each other. Some people might prefer to see more evidence of this, so here is a longer explanation:

Let $\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=n$. Recall the definition of log says that this is equivalent to \\[e^n=e^\\simplify{{p}x^2+{q}x+{r}}\\]

Therefore $n=\\simplify{{p}x^2+{q}x+{r}}$ and so we can conclude

\\[\\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=\\simplify{{p}x^2+{q}x+{r}}.\\]

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Simplifying expressions such as $b^{\\log_b(x)}$ and $b^{\\log_b(x)}$.

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