How do you solve $8 {(10)}^{x} + 2 = 26$ $8(10)^x + 2 = 26$ ?

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Answer 1 · 2018-06-24T14:10:29

$x = {log}_{10} (3) \approx 0.477$ $x=log_10(3)~~0.477$

First, rewrite to the from $a^{x} = b$ $a^x=b$ . Then use $x = {log}_{a} (b)$ $x=log_a(b)$ .

$8 {(10)}^{x} + 2 = 26$ $8(10)^x+2=26$

Subtract $2$ $2$ from both sides.

$8 {(10)}^{x} = 24$ $8(10)^x=24$

Divide both sides by $8$ $8$ .

$10^{x} = 3$ $10^x=3$

Take the common logarithm ( $log$ $log$ with base $10$ $10$ ) of both sides.

${log}_{10} (10^{x}) = {log}_{10} (3)$ $log_10(10^x)=log_10(3)$

By definition: ${log}_{a} (a^{x}) = x$ $log_a(a^x)=x$ , so:

$x = {log}_{10} (3) \approx 0.477$ $x=log_10(3)~~0.477$

How do you solve 8(10)x+2=268(10)^x + 2 = 26?