Rearrange the equation to find $x$.
\n$\\log_\\var{f}(x)=\\var{f1}$
\n$x=$ [[0]]
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\n$\\log_\\var{g1}(x)=y+\\var{g2}$
\n$x=$ [[0]]
Make $x$ the subject of the equation, leaving your answer in the form $a^{\\frac{1}{b}}$.
\n$\\log_x(y+\\var{h1})=\\var{h2}$
\n$x=$ [[0]]
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", "maxMarks": 0, "distractors": ["", "", "", "", "", ""], "minAnswers": 0, "warningType": "none", "choices": ["$\\log_a(a^x)$
", "$a^{\\log_a(x)}$
", "$e^{\\ln(x)}$
", "$\\log_{10}(x)$
", "$\\log_e(x)$
", "$\\ln(e^x)$
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\nWe can rearrange logarithms using indices.
\n\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]
\nUsing this equivalence we can rewrite $\\log_\\var{f}x=\\var{f1}$.
\n\\[\\begin{align}
x&= \\var{f}^\\var{f1} \\\\
&=\\var{f^f1}
\\end{align}\\]
\n
i)
\nWe can use the equivalence to rewrite our equation.
\n\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]
\nWe can write out our values to makes it easier.
\n\\[\\begin{align}
a&=x \\\\
b&=\\var{g1}\\\\
c&=y+\\var{g2}
\\end{align}\\]
Then we can write out our equation in the required form.
\n\\[x=\\var{g1}^{y+\\var{g2}}\\]
\n\n
We can use the same equivalence as in part b).
\n\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]
\nWe have
\n\\begin{align}
a&=y+\\var{h1} \\\\
b&=x\\\\
c&=\\var{h2}\\text{.} \\\\ \\\\
\\log_{x}(y+\\var{h1}) &= \\var{h2} \\\\
\\implies y+\\var{h1} &= x^{\\var{h2}} \\\\
x &= (y+\\var{h1})^{\\frac{1}{\\var{h2}}}
\\end{align}
The two in this list that don't equal $x$ are $\\log_e(x)$ and $\\log_{10}(x)$.
\n\\[\\begin{align}
\\log_e(x)&=\\ln(x)\\\\
\\log_{10}(x)&=\\log(x)\\text{.}
\\end{align}\\]
Changing the subject of an equation involving logarithms often requires the use of the equivalence
\n\\[\\log_ba=c \\Longleftrightarrow a=b^c\\text{.}\\]
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