How do you solve $12 = 10^{x + 5} - 7$ $12=10^(x+5)-7$ ?

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Answer 1 · 2016-09-10T10:35:19

$x = {log}_{10} (19) - 5$ $x = log_(10)(19) - 5$

We have: $12 = 10^{x + 5} - 7$ $12 = 10^(x + 5) - 7$

Let's add $7$ $7$ to both sides of the equation:

$\Rightarrow 19 = 10^{x + 5}$ $=> 19 = 10^(x + 5)$

$\Rightarrow 10^{x + 5} = 19$ $=> 10^(x + 5) = 19$

Then, let's apply ${log}_{10}$ $log_(10)$ to both sides of the equation:

$\Rightarrow {log}_{10} (10^{x + 5}) = {log}_{10} (19)$ $=> log_(10)(10^(x + 5)) = log_(10)(19)$

Using the laws of logarithms:

$\Rightarrow (x + 5) {log}_{10} (10) = {log}_{10} (19)$ $=> (x + 5) log_(10)(10) = log_(10)(19)$

$\Rightarrow (x + 5) \cdot 1 = {log}_{10} (19)$ $=> (x + 5) cdot 1 = log_(10)(19)$

$\Rightarrow x + 5 = {log}_{10} (19)$ $=> x + 5 = log_(10)(19)$

Now, to solve for $x$ $x$ , let's subtract $5$ $5$ from both sides:

$\Rightarrow x = {log}_{10} (19) - 5$ $=> x = log_(10)(19) - 5$

How do you solve 12=10x+5−712=10^(x+5)-7?