Looking for help with this equation. Trying to help boyfriends younger sister but answer is either all numbers or its impossible: $$ \log_x \left(x^5\right) = 5 $$
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No that is the question. They want to solve for x. – Scholarmate Apr 16 '19 at 21:12
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As long as $x$ is any valid log base, that will be true. – Minus One-Twelfth Apr 16 '19 at 21:12
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Let $f(x)$ be an exponential function, and $f^{-1}(x)$ be its inverse, a logarithm. Then $f^{-1}\circ f(x) = x.$ – Brevan Ellefsen Apr 16 '19 at 21:13
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2I believe the answer is for all $x\in(0,\infty)\setminus{1}$. – Stan Tendijck Apr 16 '19 at 21:26
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$\log_b a$ is really asking to which power do we need to raise $b$ to get $a$, in other words $\log_b a = x \iff b^x = a$. Therefore, $$ \log_b \left(b^5\right) = 5 $$ for all $b$ where the log is defined.
gt6989b
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Thanks I knew I wasn't going crazy. This was a question in the first section on logs before they have learned any log laws. – Scholarmate Apr 16 '19 at 21:15
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Remember the fundamental rule of $\log$s. $$\log_a(b)=c\iff a^c=b$$
It also requires $b>0$.
In other words, you seek all $b>0$ such that $b^5=b^5$. This is trivially all $b>0$
Rhys Hughes
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