How do you solve ${log}_{x} 7 = 1$ $log_x7 = 1$ ?

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Answer 1 · 2015-10-26T20:11:01

$x = 7$ $x = 7$

${log}_{a} (b) = \frac{ln (b)}{ln (a)}$ $log_a(b) =ln(b)/ln(a)$

So

${log}_{x} (7) = \frac{ln (7)}{ln (x)}$ $log_x(7) = ln(7)/ln(x)$

$\Rightarrow \frac{ln (7)}{ln (x)} = 1$ $=> ln(7)/ln(x) = 1$

$\Rightarrow ln (7) = ln (x)$ $=>ln(7) = ln(x)$

Taking exponentiel both side

$\Rightarrow x = 7$ $=>x = 7$

Answer 2 · 2015-10-26T20:20:08

x=7

Consider powers of 10
Picking one at random

$10^{2} = 100$ $10^2 = 100$

If this were to be written as log base 10 it would be:

${log}_{10} (100) = 2$ $log_10(100) = 2$

Following the same approach for your question but in you case we could reverse the process to get something we can work out.

so ${log}_{x} (7) = 1 \to x^{1} = 7$ $log_x(7) = 1 " "->" " x^(1) = 7$

Anything raised to the power of one is its own value

So $x^{1} = x = 7$ $x^1 = x = 7$

This means that if z is any number (technically you would have to say that $z \in R$ $z in R$ but I would not wary about that!)

Then ${log}_{z} (z) = 1$ $log_z(z) = 1$

How do you solve logx7=1log_x7 = 1?