How do you solve $5 (10^{y}) = 1 + 2 (10^{y})$ $5(10^y)=1+2(10^y)$ ?

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Answer 1 · 2018-05-10T17:33:14

I tried this:

Let us write it as:

$5 (10^{y}) - 2 (10^{y}) = 1$ $5(10^y)-2(10^y)=1$

collect:

$(10^{y}) [5 - 2] = 1$ $(10^y)[5-2]=1$

$(10^{y}) \cdot 3 = 1$ $(10^y)*3=1$

$10^{y} = \frac{1}{3}$ $10^y=1/3$

take the log in base $10$ $10$ of both sides:

${log}_{10} (10^{y}) = {log}_{10} (\frac{1}{3})$ $log_(10)(10^y)=log_(10)(1/3)$

giving:

$y = = {log}_{10} (\frac{1}{3}) = - 0.477121$ $y==log_(10)(1/3)=-0.477121$

How do you solve 5(10y)=1+2(10y)5(10^y)=1+2(10^y)?