How do you solve $3 (5^{x - 1}) = 21$ $3(5^(x-1))=21$ ?

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Answer 1 · 2018-07-27T10:20:38

$x = \frac{ln 35}{ln 5}$ $x=\frac{\ln35}{\ln5}$

Given that

$3 (5^{x - 1}) = 21$ $3(5^{x-1})=21$

$5^{x - 1} = \frac{21}{3}$ $5^{x-1}=21/3$

$5^{x - 1} = 7$ $5^{x-1}=7$

Taking logs on both the sides as follows

${ln 5}^{x - 1} = ln 7$ $\ln5^{x-1}=\ln 7$

$(x - 1) ln 5 = ln 7$ $(x-1)\ln 5=\ln7$

$x - 1 = \frac{ln 7}{ln 5}$ $x-1=\frac{\ln 7}{\ln 5}$

$x = \frac{ln 7}{ln 5} + 1$ $x=\frac{\ln 7}{\ln 5}+1$

$x = \frac{ln 7 + ln 5}{ln 5}$ $x=\frac{\ln 7+\ln 5}{\ln 5}$

$x = \frac{ln 35}{ln 5}$ $x=\frac{\ln35}{\ln5}$

Answer 2 · 2018-07-27T10:21:41

$x = 2.21$ $x=2.21$ (2.d.p)

$3 (5^{x - 1}) = 21$ $3(5^(x-1))=21$

$5^{x - 1} = 7$ $5^(x-1)=7$

${log}_{10} 5^{x - 1} = {log}_{10} 7$ $log_10 5^(x-1)=log_10 7$

$(x - 1) {log}_{10} 5 = {log}_{10} 7$ $(x-1) log_10 5=log_10 7$

$x - 1 = \frac{{log}_{10} 7}{{log}_{10} 5}$ $x-1=log_10 7/log_10 5$

$x = 1 + \frac{{log}_{10} 7}{{log}_{10} 5}$ $x=1+log_10 7/log_10 5$

$x = 2.21$ $x=2.21$ (2.d.p)

How do you solve 3(5x−1)=213(5^(x-1))=21?