How do you solve $2^{x + 1} = 3^{x}$ $2^(x+1) = 3^x$ ?

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How do you solve $2^{x + 1} = 3^{x}$ $2^(x+1) = 3^x$ ?

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Answer 1 · 2016-02-26T03:16:57

Convert to logarithmic form.

Explanation:

${log 2}^{x + 1} = {log 3}^{x}$ $log2^(x + 1) = log3^x$

Simplify using the rule ${log a}^{n} = n log a$ $loga^n = nloga$

$(x + 1) log 2 = x log 3$ $(x + 1)log2 = xlog3$

Distribute on the left side. Don't forget you cannot multiply directly a log with a non-log (e.g $2 \times log 3 \neq log 6$ $2 xx log3!= log6$ but is equal to $2 log 3$ $2log3$ )

$x log 2 + log 2 = x log 3$ $xlog2 + log2 = xlog3$

Put the x's to one side of the equation.

$log 2 = x log 3 - x log 2$ $log2 = xlog3 - xlog2$

$log 2 = x (log 3 - log 2)$ $log2 = x(log3 - log2)$

Simplify the right side of the equation further by using the rule $log m - log n = log (\frac{m}{n})$ $logm - logn = log(m / n)$

$log 2 = x (log (\frac{3}{2}))$ $log2 = x(log(3/2))$

$\frac{log 2}{log (\frac{3}{2})} = x$ $log2/log(3/2) = x$

You will want to ask your teacher if he/she wants the answer in exact form or rounded off. Just make sure to check.

Practice exercises:

Solve for x. Leave answers in exact form.

a) $2^{3 x} = 5^{x + 1}$ $2^(3x) = 5^(x + 1)$

b). $5^{x - 3} = 3^{2 x + 1}$ $5^(x - 3) = 3^(2x + 1)$

Challenge problem

Find the value of x in $2^{4 x - 6} = 5 \times 3^{x + 7}$ $2^(4x - 6) = 5 xx 3^(x + 7)$

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How do you solve $2^{x + 1} = 3^{x}$ $2^(x+1) = 3^x$ ?

1 Answer

Explanation:

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